Friction and Inclined Planes

Static and kinetic friction, motion on inclines with calculus analysis

Friction and Inclined Planes

Friction Forces

Static friction: fs≤μsNf_s \leq \mu_s N

Kinetic friction: fk=μkNf_k = \mu_k N

where NN is the normal force and typically μs>μk\mu_s > \mu_k.

Motion with Friction

For an object sliding with kinetic friction:

ma=Fapplied−fk=Fapplied−μkmgma = F_{applied} - f_k = F_{applied} - \mu_k mg

dvdt=Fappliedm−μkg\frac{dv}{dt} = \frac{F_{applied}}{m} - \mu_k g

Stopping Distance

Object sliding on horizontal surface with initial velocity v0v_0:

a=−μkga = -\mu_k g

vdvdx=−μkgv\frac{dv}{dx} = -\mu_k g

∫v00v dv=−μkg∫0ddx\int_{v_0}^0 v \, dv = -\mu_k g \int_0^d dx

−12v02=−μkgd-\frac{1}{2}v_0^2 = -\mu_k gd

d=v022μkgd = \frac{v_0^2}{2\mu_k g}

Inclined Planes

For angle θ\theta from horizontal:

Coordinate system: x-axis along incline (positive down), y-axis perpendicular

Weight components:

  • Parallel to incline: F∥=mgsin⁡θF_{\parallel} = mg\sin\theta
  • Perpendicular: F⊥=mgcos⁡θF_{\perp} = mg\cos\theta

Normal force: N=mgcos⁡θN = mg\cos\theta (when no vertical acceleration)

Sliding Down Incline

With kinetic friction:

ma=mgsin⁡θ−μkmgcos⁡θma = mg\sin\theta - \mu_k mg\cos\theta

a=g(sin⁡θ−μkcos⁡θ)a = g(\sin\theta - \mu_k\cos\theta)

Condition for sliding: sin⁡θ>μkcos⁡θ\sin\theta > \mu_k\cos\theta, or tan⁡θ>μk\tan\theta > \mu_k

Velocity after distance dd: v2=v02+2ad=v02+2gd(sin⁡θ−μkcos⁡θ)v^2 = v_0^2 + 2ad = v_0^2 + 2gd(\sin\theta - \mu_k\cos\theta)

Critical Angle for Static Friction

Object on the verge of sliding:

fs=μsNf_s = \mu_s N

mgsin⁡θc=μsmgcos⁡θcmg\sin\theta_c = \mu_s mg\cos\theta_c

tan⁡θc=μs\tan\theta_c = \mu_s

Motion with Time-Dependent Force

Force F(t)F(t) applied up an incline:

mdvdt=F(t)−mgsin⁡θ−μkmgcos⁡θm\frac{dv}{dt} = F(t) - mg\sin\theta - \mu_k mg\cos\theta

v(t)=v0+∫0tF(t′)m dt′−gt(sin⁡θ+μkcos⁡θ)v(t) = v_0 + \int_0^t \frac{F(t')}{m} \, dt' - gt(\sin\theta + \mu_k\cos\theta)

Example: Exponential Force

If F(t)=F0e−t/τF(t) = F_0e^{-t/\tau}:

v(t)=v0+F0τm(1−e−t/τ)−gt(sin⁡θ+μkcos⁡θ)v(t) = v_0 + \frac{F_0\tau}{m}(1 - e^{-t/\tau}) - gt(\sin\theta + \mu_k\cos\theta)

Blocks Connected on Incline

Two blocks (masses m1m_1, m2m_2) connected by rope over pulley:

Block 1 on incline at angle θ\theta, block 2 hanging vertically.

Constraint: a1=a2=aa_1 = a_2 = a (magnitude)

Block 1 (up incline positive): m1a=T−m1gsin⁡θ−fkm_1a = T - m_1g\sin\theta - f_k

Block 2 (down positive): m2a=m2g−Tm_2a = m_2g - T

Adding equations (T cancels): a=m2g−m1gsin⁡θ−μkm1gcos⁡θm1+m2a = \frac{m_2g - m_1g\sin\theta - \mu_k m_1g\cos\theta}{m_1 + m_2}

Energy Considerations with Friction

Work done by friction: Wf=−fk⋅d=−μkmgdcos⁡θW_f = -f_k \cdot d = -\mu_k mgd\cos\theta

(negative because friction opposes motion)

This work is converted to thermal energy (non-conservative force).

📚 Practice Problems

1Problem 1easy

❓ Question:

A 10 kg block is pulled up a 30° incline by a force F = 80 N parallel to the incline. The coefficient of kinetic friction is μₖ = 0.25. Find: (a) the normal force, (b) the friction force, and (c) the acceleration of the block.

💡 Show Solution

Given:

  • m = 10 kg
  • θ = 30°
  • F = 80 N (up incline)
  • μₖ = 0.25
  • g = 9.8 m/s²

(a) Normal force:

Perpendicular to incline: N=mgcos⁡θ=(10)(9.8)cos⁡(30°)N = mg\cos\theta = (10)(9.8)\cos(30°)

N=98(0.866)N = 98(0.866)

N=84.9 N\boxed{N = 84.9 \text{ N}}

(b) Friction force:

fk=μkN=(0.25)(84.9)f_k = \mu_k N = (0.25)(84.9)

fk=21.2 N\boxed{f_k = 21.2 \text{ N}} (down incline)

(c) Acceleration:

Net force along incline: Fnet=F−mgsin⁡θ−fkF_{net} = F - mg\sin\theta - f_k

Fnet=80−(10)(9.8)sin⁡(30°)−21.2F_{net} = 80 - (10)(9.8)\sin(30°) - 21.2

Fnet=80−49−21.2=9.8 NF_{net} = 80 - 49 - 21.2 = 9.8 \text{ N}

a=Fnetm=9.810a = \frac{F_{net}}{m} = \frac{9.8}{10}

a=0.98 m/s2\boxed{a = 0.98 \text{ m/s}^2} (up incline)

2Problem 2easy

❓ Question:

A 10 kg block is pulled up a 30° incline by a force F = 80 N parallel to the incline. The coefficient of kinetic friction is μₖ = 0.25. Find: (a) the normal force, (b) the friction force, and (c) the acceleration of the block.

💡 Show Solution

Given:

  • m = 10 kg
  • θ = 30°
  • F = 80 N (up incline)
  • μₖ = 0.25
  • g = 9.8 m/s²

(a) Normal force:

Perpendicular to incline: N=mgcos⁡θ=(10)(9.8)cos⁡(30°)N = mg\cos\theta = (10)(9.8)\cos(30°)

N=98(0.866)N = 98(0.866)

N=84.9 N\boxed{N = 84.9 \text{ N}}

(b) Friction force:

fk=μkN=(0.25)(84.9)f_k = \mu_k N = (0.25)(84.9)

fk=21.2 N\boxed{f_k = 21.2 \text{ N}} (down incline)

(c) Acceleration:

Net force along incline: Fnet=F−mgsin⁡θ−fkF_{net} = F - mg\sin\theta - f_k

Fnet=80−(10)(9.8)sin⁡(30°)−21.2F_{net} = 80 - (10)(9.8)\sin(30°) - 21.2

Fnet=80−49−21.2=9.8 NF_{net} = 80 - 49 - 21.2 = 9.8 \text{ N}

a=Fnetm=9.810a = \frac{F_{net}}{m} = \frac{9.8}{10}

a=0.98 m/s2\boxed{a = 0.98 \text{ m/s}^2} (up incline)

3Problem 3easy

❓ Question:

A 10 kg block is pulled up a 30° incline by a force F = 80 N parallel to the incline. The coefficient of kinetic friction is μₖ = 0.25. Find: (a) the normal force, (b) the friction force, and (c) the acceleration of the block.

💡 Show Solution

Given:

  • m = 10 kg
  • θ = 30°
  • F = 80 N (up incline)
  • μₖ = 0.25
  • g = 9.8 m/s²

(a) Normal force:

Perpendicular to incline: N=mgcos⁡θ=(10)(9.8)cos⁡(30°)N = mg\cos\theta = (10)(9.8)\cos(30°)

N=98(0.866)N = 98(0.866)

N=84.9 N\boxed{N = 84.9 \text{ N}}

(b) Friction force:

fk=μkN=(0.25)(84.9)f_k = \mu_k N = (0.25)(84.9)

fk=21.2 N\boxed{f_k = 21.2 \text{ N}} (down incline)

(c) Acceleration:

Net force along incline: Fnet=F−mgsin⁡θ−fkF_{net} = F - mg\sin\theta - f_k

Fnet=80−(10)(9.8)sin⁡(30°)−21.2F_{net} = 80 - (10)(9.8)\sin(30°) - 21.2

Fnet=80−49−21.2=9.8 NF_{net} = 80 - 49 - 21.2 = 9.8 \text{ N}

a=Fnetm=9.810a = \frac{F_{net}}{m} = \frac{9.8}{10}

a=0.98 m/s2\boxed{a = 0.98 \text{ m/s}^2} (up incline)

4Problem 4medium

❓ Question:

A block of mass m = 5.0 kg is released from rest at the top of a rough incline (θ = 20°, μₖ = 0.15) of length L = 10 m. Find: (a) the acceleration down the incline, (b) the speed at the bottom, and (c) the energy dissipated by friction.

💡 Show Solution

Given:

  • m = 5.0 kg
  • θ = 20°
  • μₖ = 0.15
  • L = 10 m
  • v₀ = 0

(a) Acceleration:

Forces along incline: ma=mgsin⁡θ−μkmgcos⁡θma = mg\sin\theta - \mu_k mg\cos\theta

a=g(sin⁡θ−μkcos⁡θ)a = g(\sin\theta - \mu_k\cos\theta)

a=9.8(sin⁡20°−0.15cos⁡20°)a = 9.8(\sin 20° - 0.15\cos 20°)

a=9.8(0.342−0.15×0.940)a = 9.8(0.342 - 0.15 \times 0.940)

a=9.8(0.342−0.141)=9.8(0.201)a = 9.8(0.342 - 0.141) = 9.8(0.201)

a=1.97 m/s2\boxed{a = 1.97 \text{ m/s}^2}

(b) Speed at bottom:

Using v2=v02+2aLv^2 = v_0^2 + 2aL:

v2=0+2(1.97)(10)=39.4v^2 = 0 + 2(1.97)(10) = 39.4

v=6.28 m/s\boxed{v = 6.28 \text{ m/s}}

(c) Energy dissipated:

Friction force: fk=μkmgcos⁡θ=(0.15)(5.0)(9.8)(0.940)=6.91 Nf_k = \mu_k mg\cos\theta = (0.15)(5.0)(9.8)(0.940) = 6.91 \text{ N}

Energy dissipated: Efriction=fk⋅L=(6.91)(10)E_{friction} = f_k \cdot L = (6.91)(10)

Efriction=69.1 J\boxed{E_{friction} = 69.1 \text{ J}}

Check with energy: Height: h=Lsin⁡θ=10(0.342)=3.42h = L\sin\theta = 10(0.342) = 3.42 m

ΔPE=mgh=(5.0)(9.8)(3.42)=168 J\Delta PE = mgh = (5.0)(9.8)(3.42) = 168 \text{ J}

KE=12mv2=12(5.0)(6.28)2=98.6 JKE = \frac{1}{2}mv^2 = \frac{1}{2}(5.0)(6.28)^2 = 98.6 \text{ J}

Efriction=ΔPE−KE=168−98.6=69.4 JE_{friction} = \Delta PE - KE = 168 - 98.6 = 69.4 \text{ J} ✓

5Problem 5medium

❓ Question:

A block of mass m = 5.0 kg is released from rest at the top of a rough incline (θ = 20°, μₖ = 0.15) of length L = 10 m. Find: (a) the acceleration down the incline, (b) the speed at the bottom, and (c) the energy dissipated by friction.

💡 Show Solution

Given:

  • m = 5.0 kg
  • θ = 20°
  • μₖ = 0.15
  • L = 10 m
  • v₀ = 0

(a) Acceleration:

Forces along incline: ma=mgsin⁡θ−μkmgcos⁡θma = mg\sin\theta - \mu_k mg\cos\theta

a=g(sin⁡θ−μkcos⁡θ)a = g(\sin\theta - \mu_k\cos\theta)

a=9.8(sin⁡20°−0.15cos⁡20°)a = 9.8(\sin 20° - 0.15\cos 20°)

a=9.8(0.342−0.15×0.940)a = 9.8(0.342 - 0.15 \times 0.940)

a=9.8(0.342−0.141)=9.8(0.201)a = 9.8(0.342 - 0.141) = 9.8(0.201)

a=1.97 m/s2\boxed{a = 1.97 \text{ m/s}^2}

(b) Speed at bottom:

Using v2=v02+2aLv^2 = v_0^2 + 2aL:

v2=0+2(1.97)(10)=39.4v^2 = 0 + 2(1.97)(10) = 39.4

v=6.28 m/s\boxed{v = 6.28 \text{ m/s}}

(c) Energy dissipated:

Friction force: fk=μkmgcos⁡θ=(0.15)(5.0)(9.8)(0.940)=6.91 Nf_k = \mu_k mg\cos\theta = (0.15)(5.0)(9.8)(0.940) = 6.91 \text{ N}

Energy dissipated: Efriction=fk⋅L=(6.91)(10)E_{friction} = f_k \cdot L = (6.91)(10)

Efriction=69.1 J\boxed{E_{friction} = 69.1 \text{ J}}

Check with energy: Height: h=Lsin⁡θ=10(0.342)=3.42h = L\sin\theta = 10(0.342) = 3.42 m

ΔPE=mgh=(5.0)(9.8)(3.42)=168 J\Delta PE = mgh = (5.0)(9.8)(3.42) = 168 \text{ J}

KE=12mv2=12(5.0)(6.28)2=98.6 JKE = \frac{1}{2}mv^2 = \frac{1}{2}(5.0)(6.28)^2 = 98.6 \text{ J}

Efriction=ΔPE−KE=168−98.6=69.4 JE_{friction} = \Delta PE - KE = 168 - 98.6 = 69.4 \text{ J} ✓

6Problem 6medium

❓ Question:

A block of mass m = 5.0 kg is released from rest at the top of a rough incline (θ = 20°, μₖ = 0.15) of length L = 10 m. Find: (a) the acceleration down the incline, (b) the speed at the bottom, and (c) the energy dissipated by friction.

💡 Show Solution

Given:

  • m = 5.0 kg
  • θ = 20°
  • μₖ = 0.15
  • L = 10 m
  • v₀ = 0

(a) Acceleration:

Forces along incline: ma=mgsin⁡θ−μkmgcos⁡θma = mg\sin\theta - \mu_k mg\cos\theta

a=g(sin⁡θ−μkcos⁡θ)a = g(\sin\theta - \mu_k\cos\theta)

a=9.8(sin⁡20°−0.15cos⁡20°)a = 9.8(\sin 20° - 0.15\cos 20°)

a=9.8(0.342−0.15×0.940)a = 9.8(0.342 - 0.15 \times 0.940)

a=9.8(0.342−0.141)=9.8(0.201)a = 9.8(0.342 - 0.141) = 9.8(0.201)

a=1.97 m/s2\boxed{a = 1.97 \text{ m/s}^2}

(b) Speed at bottom:

Using v2=v02+2aLv^2 = v_0^2 + 2aL:

v2=0+2(1.97)(10)=39.4v^2 = 0 + 2(1.97)(10) = 39.4

v=6.28 m/s\boxed{v = 6.28 \text{ m/s}}

(c) Energy dissipated:

Friction force: fk=μkmgcos⁡θ=(0.15)(5.0)(9.8)(0.940)=6.91 Nf_k = \mu_k mg\cos\theta = (0.15)(5.0)(9.8)(0.940) = 6.91 \text{ N}

Energy dissipated: Efriction=fk⋅L=(6.91)(10)E_{friction} = f_k \cdot L = (6.91)(10)

Efriction=69.1 J\boxed{E_{friction} = 69.1 \text{ J}}

Check with energy: Height: h=Lsin⁡θ=10(0.342)=3.42h = L\sin\theta = 10(0.342) = 3.42 m

ΔPE=mgh=(5.0)(9.8)(3.42)=168 J\Delta PE = mgh = (5.0)(9.8)(3.42) = 168 \text{ J}

KE=12mv2=12(5.0)(6.28)2=98.6 JKE = \frac{1}{2}mv^2 = \frac{1}{2}(5.0)(6.28)^2 = 98.6 \text{ J}

Efriction=ΔPE−KE=168−98.6=69.4 JE_{friction} = \Delta PE - KE = 168 - 98.6 = 69.4 \text{ J} ✓

7Problem 7hard

❓ Question:

A 2.0 kg block sits on a 3.0 kg wedge (incline angle θ = 30°) that rests on a frictionless table. There is no friction between block and wedge. The block is released from rest. Find: (a) the acceleration of the wedge, (b) the acceleration of the block relative to the wedge, and (c) the normal force between block and wedge.

💡 Show Solution

Given:

  • m₁ = 2.0 kg (block)
  • m₂ = 3.0 kg (wedge)
  • θ = 30°
  • All surfaces frictionless

(a) Acceleration of wedge:

Let a₂ = acceleration of wedge (to right) Let a₁ = acceleration of block relative to wedge (down incline)

For block, horizontal components: m1a1x=Nsin⁡θm_1 a_{1x} = N\sin\theta

where a1x=a2−a1cos⁡θa_{1x} = a_2 - a_1\cos\theta (acceleration of block in lab frame)

For block, vertical components: 0=Ncos⁡θ−m1g0 = N\cos\theta - m_1 g

From second equation: N=m1gcos⁡θN = \frac{m_1 g}{\cos\theta}

For wedge: m2a2=−Nsin⁡θm_2 a_2 = -N\sin\theta

Combining: m1(a2−a1cos⁡θ)=m1gsin⁡θcos⁡θm_1(a_2 - a_1\cos\theta) = \frac{m_1 g\sin\theta}{\cos\theta}

m2a2=−m1gsin⁡θcos⁡θm_2 a_2 = -\frac{m_1 g\sin\theta}{\cos\theta}

From wedge equation: a2=−m1gtan⁡θm2=−(2.0)(9.8)tan⁡30°3.0a_2 = -\frac{m_1 g\tan\theta}{m_2} = -\frac{(2.0)(9.8)\tan 30°}{3.0}

a2=−19.6(0.577)3.0=−3.77 m/s2a_2 = -\frac{19.6(0.577)}{3.0} = -3.77 \text{ m/s}^2

a2=3.77 m/s2 (to left)\boxed{a_2 = 3.77 \text{ m/s}^2 \text{ (to left)}}

(b) Acceleration relative to wedge:

From block equation: a2−a1cos⁡θ=gtan⁡θa_2 - a_1\cos\theta = g\tan\theta

−3.77−a1(0.866)=(9.8)(0.577)=5.65-3.77 - a_1(0.866) = (9.8)(0.577) = 5.65

a1=−3.77−5.650.866=−10.9 m/s2a_1 = \frac{-3.77 - 5.65}{0.866} = -10.9 \text{ m/s}^2

a1=10.9 m/s2 (down incline)\boxed{a_1 = 10.9 \text{ m/s}^2 \text{ (down incline)}}

(c) Normal force:

N=m1gcos⁡θ=(2.0)(9.8)cos⁡30°N = \frac{m_1 g}{\cos\theta} = \frac{(2.0)(9.8)}{\cos 30°}

N=19.60.866N = \frac{19.6}{0.866}

N=22.6 N\boxed{N = 22.6 \text{ N}}

8Problem 8hard

❓ Question:

A 2.0 kg block sits on a 3.0 kg wedge (incline angle θ = 30°) that rests on a frictionless table. There is no friction between block and wedge. The block is released from rest. Find: (a) the acceleration of the wedge, (b) the acceleration of the block relative to the wedge, and (c) the normal force between block and wedge.

💡 Show Solution

Given:

  • m₁ = 2.0 kg (block)
  • m₂ = 3.0 kg (wedge)
  • θ = 30°
  • All surfaces frictionless

(a) Acceleration of wedge:

Let a₂ = acceleration of wedge (to right) Let a₁ = acceleration of block relative to wedge (down incline)

For block, horizontal components: m1a1x=Nsin⁡θm_1 a_{1x} = N\sin\theta

where a1x=a2−a1cos⁡θa_{1x} = a_2 - a_1\cos\theta (acceleration of block in lab frame)

For block, vertical components: 0=Ncos⁡θ−m1g0 = N\cos\theta - m_1 g

From second equation: N=m1gcos⁡θN = \frac{m_1 g}{\cos\theta}

For wedge: m2a2=−Nsin⁡θm_2 a_2 = -N\sin\theta

Combining: m1(a2−a1cos⁡θ)=m1gsin⁡θcos⁡θm_1(a_2 - a_1\cos\theta) = \frac{m_1 g\sin\theta}{\cos\theta}

m2a2=−m1gsin⁡θcos⁡θm_2 a_2 = -\frac{m_1 g\sin\theta}{\cos\theta}

From wedge equation: a2=−m1gtan⁡θm2=−(2.0)(9.8)tan⁡30°3.0a_2 = -\frac{m_1 g\tan\theta}{m_2} = -\frac{(2.0)(9.8)\tan 30°}{3.0}

a2=−19.6(0.577)3.0=−3.77 m/s2a_2 = -\frac{19.6(0.577)}{3.0} = -3.77 \text{ m/s}^2

a2=3.77 m/s2 (to left)\boxed{a_2 = 3.77 \text{ m/s}^2 \text{ (to left)}}

(b) Acceleration relative to wedge:

From block equation: a2−a1cos⁡θ=gtan⁡θa_2 - a_1\cos\theta = g\tan\theta

−3.77−a1(0.866)=(9.8)(0.577)=5.65-3.77 - a_1(0.866) = (9.8)(0.577) = 5.65

a1=−3.77−5.650.866=−10.9 m/s2a_1 = \frac{-3.77 - 5.65}{0.866} = -10.9 \text{ m/s}^2

a1=10.9 m/s2 (down incline)\boxed{a_1 = 10.9 \text{ m/s}^2 \text{ (down incline)}}

(c) Normal force:

N=m1gcos⁡θ=(2.0)(9.8)cos⁡30°N = \frac{m_1 g}{\cos\theta} = \frac{(2.0)(9.8)}{\cos 30°}

N=19.60.866N = \frac{19.6}{0.866}

N=22.6 N\boxed{N = 22.6 \text{ N}}

9Problem 9hard

❓ Question:

A 2.0 kg block sits on a 3.0 kg wedge (incline angle θ = 30°) that rests on a frictionless table. There is no friction between block and wedge. The block is released from rest. Find: (a) the acceleration of the wedge, (b) the acceleration of the block relative to the wedge, and (c) the normal force between block and wedge.

💡 Show Solution

Given:

  • m₁ = 2.0 kg (block)
  • m₂ = 3.0 kg (wedge)
  • θ = 30°
  • All surfaces frictionless

(a) Acceleration of wedge:

Let a₂ = acceleration of wedge (to right) Let a₁ = acceleration of block relative to wedge (down incline)

For block, horizontal components: m1a1x=Nsin⁡θm_1 a_{1x} = N\sin\theta

where a1x=a2−a1cos⁡θa_{1x} = a_2 - a_1\cos\theta (acceleration of block in lab frame)

For block, vertical components: 0=Ncos⁡θ−m1g0 = N\cos\theta - m_1 g

From second equation: N=m1gcos⁡θN = \frac{m_1 g}{\cos\theta}

For wedge: m2a2=−Nsin⁡θm_2 a_2 = -N\sin\theta

Combining: m1(a2−a1cos⁡θ)=m1gsin⁡θcos⁡θm_1(a_2 - a_1\cos\theta) = \frac{m_1 g\sin\theta}{\cos\theta}

m2a2=−m1gsin⁡θcos⁡θm_2 a_2 = -\frac{m_1 g\sin\theta}{\cos\theta}

From wedge equation: a2=−m1gtan⁡θm2=−(2.0)(9.8)tan⁡30°3.0a_2 = -\frac{m_1 g\tan\theta}{m_2} = -\frac{(2.0)(9.8)\tan 30°}{3.0}

a2=−19.6(0.577)3.0=−3.77 m/s2a_2 = -\frac{19.6(0.577)}{3.0} = -3.77 \text{ m/s}^2

a2=3.77 m/s2 (to left)\boxed{a_2 = 3.77 \text{ m/s}^2 \text{ (to left)}}

(b) Acceleration relative to wedge:

From block equation: a2−a1cos⁡θ=gtan⁡θa_2 - a_1\cos\theta = g\tan\theta

−3.77−a1(0.866)=(9.8)(0.577)=5.65-3.77 - a_1(0.866) = (9.8)(0.577) = 5.65

a1=−3.77−5.650.866=−10.9 m/s2a_1 = \frac{-3.77 - 5.65}{0.866} = -10.9 \text{ m/s}^2

a1=10.9 m/s2 (down incline)\boxed{a_1 = 10.9 \text{ m/s}^2 \text{ (down incline)}}

(c) Normal force:

N=m1gcos⁡θ=(2.0)(9.8)cos⁡30°N = \frac{m_1 g}{\cos\theta} = \frac{(2.0)(9.8)}{\cos 30°}

N=19.60.866N = \frac{19.6}{0.866}

N=22.6 N\boxed{N = 22.6 \text{ N}}

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