Question

A box of mass $m = 2$ kg rests on a rough horizontal surface where coefficients of static and kinetic frictions are $\mu_s = \mu_k = 0.25$. The box is acted upon by a horizontal force $F$, which varies with time as shown in the given graph.

Answer 1

a) 11.25 m/s, b) 8.25 s

Answer 2

The problem involves analyzing the motion of a box on a rough horizontal surface under the influence of a time-varying horizontal force.

Given data:

• Mass of the box, $m = 2$ kg
• Coefficients of static and kinetic friction, $\mu_s = \mu_k = 0.25$
• Acceleration due to gravity, $g = 10 \text{ m/s}^2$ (assumed for simpler calculations, common in such problems unless specified otherwise).

1. Calculate the friction force: The normal force on the box is $N = mg = 2 \text{ kg} \times 10 \text{ m/s}^2 = 20$ N. The maximum static friction force is $f_{s,max} = \mu_s N = 0.25 \times 20 \text{ N} = 5$ N. The kinetic friction force is $f_k = \mu_k N = 0.25 \times 20 \text{ N} = 5$ N. Since $\mu_s = \mu_k$, the friction force opposing motion will be constant at 5 N once the box starts moving.

2. Analyze the applied force $F(t)$ from the graph:

• For $0 \le t \le 2$ s, the force increases linearly from 0 to 20 N. So, $F(t) = \frac{20}{2}t = 10t$ N.
• For $2 \le t \le 4$ s, the force decreases linearly from 20 N to 0 N. So, $F(t) = 20 - \frac{20}{2}(t-2) = 20 - 10(t-2) = 20 - 10t + 20 = 40 - 10t$ N.
• For $t > 4$ s, the applied force $F(t) = 0$ N.

Part (a): Determine the maximum velocity acquired by the box.

• When does the box start moving? The box starts moving when the applied force $F(t)$ exceeds the maximum static friction force $f_{s,max}$. $10t = 5 \Rightarrow t = 0.5$ s. So, the box starts moving at $t_1 = 0.5$ s with initial velocity $v(0.5) = 0$.

• When does the maximum velocity occur? The maximum velocity occurs when the net force on the box becomes zero, i.e., when the applied force $F(t)$ equals the kinetic friction force $f_k$. After this point, $F(t)$ becomes less than $f_k$, and the box starts decelerating. We need to find $t$ such that $F(t) = 5$ N. For $2 \le t \le 4$ s: $40 - 10t = 5 \Rightarrow 10t = 35 \Rightarrow t = 3.5$ s. So, the maximum velocity is acquired at $t_{max\_v} = 3.5$ s.

• Calculate the maximum velocity using the impulse-momentum theorem: The change in momentum is equal to the net impulse: $m\Delta v = \int F_{net}(t) dt$. Here, $F_{net}(t) = F(t) - f_k = F(t) - 5$. $m(v_{max} - v(0.5)) = \int_{0.5}^{3.5} (F(t) - 5) dt$. Since $v(0.5) = 0$: $2 \times v_{max} = \int_{0.5}^{3.5} (F(t) - 5) dt$. We split the integral into two parts based on the definition of $F(t)$: $2 \times v_{max} = \int_{0.5}^{2} (10t - 5) dt + \int_{2}^{3.5} (40 - 10t - 5) dt$ $2 \times v_{max} = \int_{0.5}^{2} (10t - 5) dt + \int_{2}^{3.5} (35 - 10t) dt$

  Evaluate the first integral: $\left[ 5t^2 - 5t \right]_{0.5}^{2} = (5(2)^2 - 5(2)) - (5(0.5)^2 - 5(0.5))$ $= (20 - 10) - (1.25 - 2.5) = 10 - (-1.25) = 11.25$ Ns.

  Evaluate the second integral: $\left[ 35t - 5t^2 \right]_{2}^{3.5} = (35(3.5) - 5(3.5)^2) - (35(2) - 5(2)^2)$ $= (122.5 - 5 \times 12.25) - (70 - 20)$ $= (122.5 - 61.25) - 50 = 61.25 - 50 = 11.25$ Ns.

  Total impulse = $11.25 + 11.25 = 22.5$ Ns. So, $2 \times v_{max} = 22.5$ $v_{max} = \frac{22.5}{2} = 11.25$ m/s.

Part (b): Determine the instant the box comes to rest.

• Velocity at $t=4$ s: The box starts decelerating after $t=3.5$ s. We need to find its velocity at $t=4$ s, after which the applied force becomes zero. From $t=3.5$ s to $t=4$ s, $F(t) = 40 - 10t$. The net force is $F_{net}(t) = F(t) - f_k = (40 - 10t) - 5 = 35 - 10t$. The acceleration is $a(t) = \frac{F_{net}(t)}{m} = \frac{35 - 10t}{2} = 17.5 - 5t$. $v(4) = v(3.5) + \int_{3.5}^{4} a(t) dt$ $v(4) = 11.25 + \int_{3.5}^{4} (17.5 - 5t) dt$ $v(4) = 11.25 + \left[ 17.5t - \frac{5t^2}{2} \right]_{3.5}^{4}$ $v(4) = 11.25 + \left( (17.5 \times 4 - \frac{5 \times 4^2}{2}) - (17.5 \times 3.5 - \frac{5 \times 3.5^2}{2}) \right)$ $v(4) = 11.25 + \left( (70 - 40) - (61.25 - 30.625) \right)$ $v(4) = 11.25 + (30 - 30.625) = 11.25 - 0.625 = 10.625$ m/s.

• Time to rest after $t=4$ s: For $t > 4$ s, the applied force $F(t) = 0$. The only horizontal force acting on the box is the kinetic friction force $f_k = 5$ N, acting opposite to the motion. The net force is $F_{net} = -f_k = -5$ N. The acceleration is $a = \frac{-5}{2} = -2.5$ m/s$^2$ (constant deceleration). Let $t'$ be the additional time taken for the box to come to rest from $t=4$ s. Using the equation $v = v_0 + at'$: $0 = v(4) + a t'$ $0 = 10.625 + (-2.5)t'$ $2.5t' = 10.625$ $t' = \frac{10.625}{2.5} = 4.25$ s.

• Instant the box comes to rest: The total time from the start is $t_{rest} = 4 \text{ s} + t' = 4 + 4.25 = 8.25$ s. At $t=8.25$ s, the applied force is $F=0$. Since $F < f_{s,max}$ (0 N < 5 N), the box will remain at rest.

The final answer is $\boxed{\text{a) 11.25 m/s, b) 8.25 s}}$

Explanation of the solution:

1. Friction Calculation: The constant kinetic friction force is calculated as $f_k = \mu_k mg = 0.25 \times 2 \times 10 = 5$ N. The box starts moving when the applied force $F(t)$ exceeds this value.
2. Maximum Velocity (Part a):
   • The box starts moving at $t=0.5$ s (when $F(t) = 10t = 5$ N).
   • Maximum velocity occurs when the net force is zero, meaning $F(t) = f_k = 5$ N during the deceleration phase of the applied force. This happens at $t=3.5$ s (when $F(t) = 40 - 10t = 5$ N).
   • The maximum velocity is found by calculating the total impulse (area under the net force vs. time graph, $F_{net}(t) = F(t) - f_k$) from $t=0.5$ s to $t=3.5$ s and dividing by the mass.
   • Impulse = $\int_{0.5}^{3.5} (F(t) - 5) dt = \int_{0.5}^{2} (10t - 5) dt + \int_{2}^{3.5} (35 - 10t) dt = 11.25 \text{ Ns} + 11.25 \text{ Ns} = 22.5 \text{ Ns}$.
   • Maximum velocity $v_{max} = \frac{\text{Impulse}}{m} = \frac{22.5}{2} = 11.25$ m/s.
3. Time to Rest (Part b):
   • First, calculate the velocity at $t=4$ s, when the applied force becomes zero. This is done by integrating the acceleration from $t=3.5$ s to $t=4$ s. $v(4) = v(3.5) + \int_{3.5}^{4} (\frac{35 - 10t}{2}) dt = 11.25 - 0.625 = 10.625$ m/s.
   • After $t=4$ s, the applied force is zero, so the box decelerates only due to kinetic friction. The acceleration is constant: $a = \frac{-f_k}{m} = \frac{-5}{2} = -2.5$ m/s$^2$.
   • The additional time $t'$ to stop from $10.625$ m/s with this deceleration is $t' = \frac{0 - 10.625}{-2.5} = 4.25$ s.
   • The total time when the box comes to rest is $t_{rest} = 4 \text{ s} + 4.25 \text{ s} = 8.25$ s.

Answer: (a) The maximum velocity acquired by the box is 11.25 m/s. (b) The instant the box comes to rest is 8.25 s.