Question

A block of mass 2 kg slides on a rough horizontal surface with initial speed $10~ms^{-1}$. The distance travelled by the block before it stops is ($g=10~ms^{-2}$, $\mu_{k}=0.2$)

• A. 5 m
• B. 12 m
• C. 15 m
• D. 21 m
• E. 25 m

Hint:

Using the Work-Energy Theorem is even faster! Work done by friction = Change in Kinetic Energy. $\mu_k mgs = \frac{1}{2}mu^2$. Canceling $m$ and solving for $s$ gives $s = \frac{u^2}{2\mu_k g}$.

Answer 1

Answer: (E)

Concept:
Physics - Work-Energy Theorem or Kinematics with Friction.
Step 1: Find the frictional force and acceleration.
The kinetic frictional force is $f_k = \mu_k N = \mu_k mg$. The acceleration (deceleration) $a$ produced by friction is: $$ a = -\frac{f_k}{m} = -\mu_k g $$ Substitute the values $\mu_k = 0.2$ and $g = 10$: $$ a = -(0.2)(10) = -2~ms^{-2} $$
Step 2: Identify the kinematic variables.

• Initial velocity ($u$) = $10~ms^{-1}$
• Final velocity ($v$) = $0~ms^{-1}$ (the block stops)
• Acceleration ($a$) = $-2~ms^{-2}$

Step 3: Apply the third equation of motion.
$$ v^2 = u^2 + 2as $$ $$ 0^2 = (10)^2 + 2(-2)s $$
Step 4: Solve for the distance (s).
$$ 0 = 100 - 4s $$ $$ 4s = 100 $$ $$ s = 25 \text{ m} $$