Question

A 1 kg block moving on horizontal rough surface of friction coefficient 0.6 is pushed with a force varying with time t in seconds as shown in figure. If initial velocity of block was 4.5 m/s. Find the velocity (in m/s) at t = 3 s.

Answer 1

0

Answer 2

The problem asks us to find the velocity of a 1 kg block at t = 3 s, given its initial velocity, the coefficient of friction, and a time-varying applied force.

1. Determine the applied force F(t): The given graph shows the applied force F (in N) varying with time t (in s). It's a straight line passing through points (0, 9) and (3, 0). The slope of the line is $m = \frac{0 - 9}{3 - 0} = \frac{-9}{3} = -3$ N/s. Using the point-slope form $F - F_1 = m(t - t_1)$ with $(t_1, F_1) = (0, 9)$: $F(t) - 9 = -3(t - 0)$ $F(t) = 9 - 3t$

2. Calculate the kinetic frictional force (f_k): The block is on a horizontal surface. The normal force (N) is equal to the gravitational force (mg). Given mass (m) = 1 kg. We assume $g = 10 \text{ m/s}^2$ for simplicity, which is common in such problems unless specified otherwise. $N = mg = 1 \text{ kg} \times 10 \text{ m/s}^2 = 10 \text{ N}$. The coefficient of friction (μ) = 0.6. The kinetic frictional force is $f_k = \mu N = 0.6 \times 10 \text{ N} = 6 \text{ N}$. Since the block's initial velocity is positive (4.5 m/s), the friction force opposes this motion, acting in the negative direction.

3. Apply Newton's Second Law: The net force ($F_{net}$) on the block is the applied force minus the frictional force, as long as the block is moving. $F_{net} = F(t) - f_k$ $F_{net} = (9 - 3t) - 6$ $F_{net} = 3 - 3t$

   According to Newton's Second Law, $F_{net} = ma$, where 'a' is the acceleration. $ma = 3 - 3t$ Given $m = 1 \text{ kg}$: $1 \times a = 3 - 3t$ $a(t) = 3 - 3t$

4. Integrate acceleration to find velocity: Acceleration is the rate of change of velocity, $a = \frac{dv}{dt}$. $\frac{dv}{dt} = 3 - 3t$ To find the velocity at $t = 3 \text{ s}$, we integrate this equation from $t = 0$ to $t = 3 \text{ s}$. The initial velocity (u) at $t = 0$ is $4.5 \text{ m/s}$. Let the final velocity at $t = 3 \text{ s}$ be $v$. $\int_{u}^{v} dv = \int_{0}^{3} (3 - 3t) dt$ $[v]_{u}^{v} = [3t - \frac{3t^2}{2}]_{0}^{3}$ $v - u = (3 \times 3 - \frac{3 \times 3^2}{2}) - (3 \times 0 - \frac{3 \times 0^2}{2})$ $v - u = (9 - \frac{27}{2}) - 0$ $v - u = 9 - 13.5$ $v - u = -4.5 \text{ m/s}$

5. Calculate the final velocity: Substitute the initial velocity $u = 4.5 \text{ m/s}$: $v - 4.5 = -4.5$ $v = 4.5 - 4.5$ $v = 0 \text{ m/s}$

   The block comes to rest at $t = 3 \text{ s}$. Since the velocity becomes exactly zero, it does not reverse its direction of motion. At $t=3s$, the applied force is $F(3)=0$. The maximum static friction is $f_{s,max} = \mu_s N$. Assuming $\mu_s = \mu_k = 0.6$, $f_{s,max}=6N$. Since $F(3)=0 < 6N$, the block will remain at rest after $t=3s$.