A man pushing a crate of mass

Question

m = 92.0 kg
at a speed of
v = 0.875 m/s
encounters a rough horizontal surface of length
ℓ = 0.65 m
as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.359 and he exerts a constant horizontal force of 298 N on the crate.

A man pushes a crate labeled m, which moves with a velocity vector v to the right, on a horizontal surface. The horizontal surface is textured from the right edge of the crate to a horizontal distance ℓ from the right edge of the crate.
(a) Find the magnitude and direction of the net force on the crate while it is on the rough surface.
magnitude
N
direction
---Select---

(b) Find the net work done on the crate while it is on the rough surface.
J

(c) Find the speed of the crate when it reaches the end of the rough surface.
m/s

Answer 1

To solve this problem, we need to analyze the forces acting on the crate while it's on the rough surface, as well as calculate the work done and the final speed.

(a) Finding the magnitude and direction of the net force on the crate.

1. Calculate the force of friction (F_friction): \[ F_{\text{friction}} = \mu_k \cdot N \] where \( \mu_k = 0.359 \) is the coefficient of kinetic friction, and \( N \) (the normal force) equals the weight of the crate: \[ N = m \cdot g = 92.0 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 902.52 , \text{N} \] \[ F_{\text{friction}} = 0.359 \cdot 902.52 = 323.32 , \text{N} \]

2. Calculate the net force (F_net): The net force can be calculated by subtracting the force of friction from the applied force: \[ F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 298 , \text{N} - 323.32 , \text{N} = -25.32 , \text{N} \] The negative sign indicates that the net force is directed opposite to the direction of motion.

Magnitude: \[ 25.32 , \text{N} \]

Direction: Opposite to the direction of motion (to the left).

(b) Finding the net work done on the crate while it is on the rough surface.

Work done is calculated using the formula: \[ W_{\text{net}} = F_{\text{net}} \cdot d \] where \( d = 0.65 , \text{m} \): \[ W_{\text{net}} = -25.32 , \text{N} \cdot 0.65 , \text{m} = -16.47 , \text{J} \] The negative sign indicates that the work done by friction is against the motion of the crate.

Net Work Done: \[ -16.47 , \text{J} \]

(c) Finding the speed of the crate when it reaches the end of the rough surface.

To find the final speed of the crate when it reaches the end of the rough surface, we use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.

1. Initial kinetic energy (KE_initial): \[ KE_{\text{initial}} = \frac{1}{2} m v^2 = \frac{1}{2} \cdot 92.0 , \text{kg} \cdot (0.875 , \text{m/s})^2 = 35.40 , \text{J} \]

2. Final kinetic energy (KE_final): The work done by the net force is equal to the change in kinetic energy: \[ W_{\text{net}} = KE_{\text{final}} - KE_{\text{initial}} \] \[ -16.47 , \text{J} = KE_{\text{final}} - 35.40 , \text{J} \] \[ KE_{\text{final}} = 35.40 , \text{J} - 16.47 , \text{J} = 18.93 , \text{J} \]

3. Final speed (v_final): \[ KE_{\text{final}} = \frac{1}{2} m v_{\text{final}}^2 \] \[ 18.93 , \text{J} = \frac{1}{2} \cdot 92.0 , \text{kg} \cdot v_{\text{final}}^2 \] \[ v_{\text{final}}^2 = \frac{2 \cdot 18.93}{92.0} = \frac{37.86}{92.0} \approx 0.411 \] \[ v_{\text{final}} \approx \sqrt{0.411} \approx 0.640 , \text{m/s} \]

Final Speed: \[ 0.640 , \text{m/s} \]

Summary of Answers:

(a) Magnitude: 25.32 N; Direction: Opposite to the direction of motion.

(b) Net Work Done: -16.47 J.

(c) Speed when reaching the end of the rough surface: 0.640 m/s.