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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Answer:
a) Ep = 5886[J]; b) v = 14[m/s]; c) W = 5886[J]; d) F = 1763.4[N]
Explanation:
a)
The potential energy can be found using the following expression, we will take the ground level as the reference point where the potential energy is equal to zero.
b)
Since energy is conserved, that is, potential energy is transformed into kinetic energy, the moment the harpsichord touches water, all potential energy is transformed into kinetic energy.
c)
The work is equal to
W = 5886 [J]
d)
We need to use the following equation and find the deceleration of the diver at the moment when he stops his velocity is zero.
By performing a sum of forces equal to the product of mass by acceleration (newton's second law), we can find the force that acts to reduce the speed of the diver to zero.
m*g - F = m*a
F = m*a - m*g
F = (60*39.2) - (60*9.81)
F = 1763.4 [N]