Question

A truck of mass 8 ton is carrying a block of mass 2 ton. If a breaking force of 25 kN is applied on the truck, then the frictional force acting on the block is (Coefficient of static friction between the block and the truck is 0.3)

• A. 6250 N
• B. 6000 N
• C. 5000 N
• D. 1000 N

Hint:

In problems with friction between two moving objects, first calculate the acceleration of the system as if they were a single object. Then, isolate the top object and determine the frictional force required to produce that acceleration. Finally, compare this required force to the maximum static friction to see if slipping occurs.

Answer 1

The Correct Option is C

First, let's find the deceleration of the entire system (truck + block) assuming they move together.
Total mass $M_{total} = M_{truck} + M_{block} = 8 \text{ ton} + 2 \text{ ton} = 10 \text{ ton} = 10000$ kg.
Breaking force $F_{breaking} = 25$ kN = 25000 N.
Using Newton's second law, $F = ma$, the deceleration of the system is:
$a = \frac{F_{breaking}}{M_{total}} = \frac{25000 \text{ N}}{10000 \text{ kg}} = 2.5 \text{ m/s}^2$.
Now, consider the block of mass $m = 2$ ton = 2000 kg.
The only horizontal force acting on the block is the force of static friction, $f_s$, exerted by the truck's surface. This frictional force is what causes the block to decelerate along with the truck.
The force required to decelerate the block at a rate of $2.5 \text{ m/s}^2$ is:
$F_{required} = m \cdot a = (2000 \text{ kg})(2.5 \text{ m/s}^2) = 5000$ N.
This required force must be provided by static friction. So, the acting frictional force is $f_s = 5000$ N.
We must check if this required frictional force is less than or equal to the maximum possible static friction, $f_{s,max}$.
$f_{s,max} = \mu_s \cdot N = \mu_s \cdot (mg)$.
$f_{s,max} = 0.3 \times (2000 \text{ kg}) \times (10 \text{ m/s}^2) = 6000$ N.
Since the required force (5000 N) is less than the maximum available static friction (6000 N), the block does not slip and moves together with the truck.
Therefore, the frictional force acting on the block is the required force, which is 5000 N.