The velocity of 2 kg body is changed from `(4hat(i)+3hat(j)) ms^(-1)` to `6 ms^(-1)`. The work done on the body is

The velocity of a 2 kg body is changed from (4hati +3hatj) ms^(-1) to 6hat ms^(-1) . The work done on the body is

A particle velocity changes from (2hat(i)+3hat(j))ms^(-1) to (2hat(i)-3hat(j))ms^(-1) in 2 s. The acceleration in ms^(-2) is

A particle velocity changes from (2hat(i)+3hat(j))ms^(-1) to (2hat(i)-3hat(j))ms^(-1) in 2 s. The acceleration in ms^(-2) is

A force (4hat(i)+hat(j)-2hat(k)) ms^(-1) The power exerted is

The angular velocity of a rotating body is vec omega = 4 hat i + hat j - 2 hat k . The linear velocity of the body whose position vector 2hati + 3 hat j - 3 hat k is

A force of (2.6 hat(i) + 1.6 hat(j)) N acts on a body of mass 2 kg. If the velocity of the body at time, t = 0 is (3.6 hat(i) - 4.8 hat(j)) ms^(-1) , the time at which the body will just have a velocity along x-axis only is

A force (2hat I + hat j - hat k)N acts on a body which is initially at rest. At the end of 20 sec the velocity of the body is (4 hat i + 2 hat j – 2 hat k) ms^(-1) ,then mass of the body is

A force is applied on a body of mass 20 kg moving with a velocity of 40ms^(-1) . The body attains a velocity of 50ms^(-1) in 2 second. Calculate the work done by the body.