Question

A vehicle of mass 1000 kg is moving with a velocity of 15$m{s^{( - 1)}}$ It is brought to rest by applying brakes and locking the wheels. If the sliding friction between the tyres and the road is 6000 N, then the distance moved by the vehicle before coming to rest is_________
A. 37.5 m
B. 18.75 m
C. 75 m
D. 15 m

Answer 1

Friction is the force that prevents solid surfaces, fluid layers, and material components sliding against each other from moving in the same direction. Sliding is a form of frictional motion that occurs when two surfaces come into touch with one other. Rolling motion is the polar opposite of this. Bearings can experience both forms of motion.

Complete step by step solution:
Sliding friction (also known as kinetic friction) is a contact force that prevents two objects or an item and a surface from sliding. Sliding friction is nearly always lower than static friction, which is why it is simpler to move an item once it has started moving than getting it to start moving from a stop. Work is the energy delivered to or from an item by applying force along a displacement in physics. It is frequently expressed as the product of force and displacement in its simplest form.
The product is the work W done by a constant force of size F on a point moving a displacement s in a straight line in the force's direction.
$W = F.s$
Work on a free (no fields), rigid (no internal degrees of freedom) body is equal to the change in kinetic energy${E_k}$ corresponding to the linear and angular velocity of that body, according to Newton's second law.
$W = \Delta {E_k}$
$W = \dfrac{1}{2}m{v^2}$
Suppose we equate both the equations we get
$F.s = \dfrac{1}{2}m{v^2}$
$s = \dfrac{{\dfrac{1}{2}m{v^2}}}{F}$
Hence we obtain from the question
$s = \dfrac{{\dfrac{1}{2}(1000)({{15}^2})}}{{6000}}$
$\Rightarrow s = 18.75m$
Hence option B is correct.

Note:
The object's displacement in the system is determined by constraint forces, which confine it to a certain range. In the case of a slope with gravity, for example, the object is stuck to the slope and, when tied to a taut string, it cannot move outwards to make the string tauter. It removes all displacements in that direction, i.e., the velocity in the constraint's direction is set to zero, preventing the constraint forces from doing work on the system.