The mass of a car is 1200 kg. It comes to rest due to retardation of ` 2 ms^(-2)` .
What would be the force (frictional force) acting between the car and road ?

When F = 2N, the frictional force between 5 kg block and ground is

When F = 2n, the frictional force between 10 kg block 5 kg block is

A man of mass 80kg pushes box of mass 20kg horizontaly. The man moves the box with a constant acceleration of 2 m//s^(2) but his foot does not slip on the ground. There is no friction between the box and the ground. There is no friction between the box and the ground. whereas there is sufficient friction between the man's foot and the ground to prevent him from slipping. Assertion:- The force applied by the man on the box is equal and opposite to the force applied by the force applid by the box on the man. Reason:- Friction force applied by the ground on the man is 200N.

A car of mass 1000kg moves from point A to B . If kinetic energy of car at point A is 100KJ and at B is 220 Kj . Then find work done by friction force on the car ?

On complete combustion a litre of petrol gives off heat equivalent to 3 xx 10^(7) J. In a test drive, a car weighing 1200 kg including the mass of driver, runs 15 km per litre while moving with a uniform speed on a straight track. Assuming that friction offered by the road surface and air to be uniform. calculate the force of friction acting on the car during the test drive. if the efficiency of the car engine were 0.5.

Certain force acting on a 20 kg mass changes its velocity from 5ms^(-1) to 2ms^(-1) . Calculate the work done by the force.