A 17.5-kg block is dragged over a rough, horizontal surface by a 75-N force acting at 21° above the horizontal. The block is dis
placed 5.7 m, and the coefficient of kinetic friction is 0.150.
Requried:
a. Find the work done on the block by the 75-N force.
b. Find the work done on the block by the normal force.
c. Find the work done on the block by the gravitational force.
d. What is the increase in internal energy of the block-surface system due to friction?
e. Find the total change in the block's kinetic energy.
Requried:
a. Find the work done on the block by the 75-N force.
b. Find the work done on the block by the normal force.
c. Find the work done on the block by the gravitational force.
d. What is the increase in internal energy of the block-surface system due to friction?
e. Find the total change in the block's kinetic energy.
1 answer:
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Let both the balls have the same mass equals to m.
Let and
be the speed of the ball1 and the ball2 respectively, such that
Assuming that both the balls are at the same level with respect to the ground, so let h be the height from the ground.
The total energy of ball1= Kinetic energy of ball1 + Potential energy of ball1. The Kinetic energy of any object moving with speed, , is
and the potential energy is due to the change in height is [where
is the acceleration due to gravity]
So, the total energy of ball1,
and the total energy of ball1,
.
Here, the potential energy for both the balls are the same, but the kinetic energy of the ball1 is higher the ball2 as the ball1 have the higher speed, refer equation (i)
So,
Now, from equations (ii) and (iii)
The total energy of ball1 hi higher than the total energy of ball2.