Answer:
(a) 8.117 (b) 0.742
Explanation:
We have given distance s =40 m time t=7.5 sec
Final velocity v =2.55 m/sec
From the first equation of motion (negative sign because there is retrdation as the truck speed is slowing down )
So --------------eqn 1
From the second equation of motion ( negative sign because there is retrdation as the truck speed is slowing down )
So
------------------eqn 2
On solving eqn1 and eqn 2
u=8.117 m/sec and a=-0.742
Answer:
2.83 m/s
Explanation:
Given:
Mass of the cart (m) = 20.0 kg
Initial velocity of the cart (u) = 0 m/s
Final velocity (v) = ? m/s
Displacement of the cart (S) = 8.0 m
Horizontal force acting on the cart (F) = 10.0 N
Surface is frictionless. So, only horizontal force is the force acting on the cart.
Now, as per work-energy theorem, the work done by the net force acting on a body is equal to the change in the kinetic energy of the body.
Here, the work done by the horizontal force is given as:
The change in kinetic energy is given as:
Now, from work-energy theorem:
Therefore, the speed of the cart when it has been pushed 8.0 m is 2.83 m/s.