crate is moved across the floor by pulling a
rope tied to it. The force on the crate is of
magnitude 450 N and is directed at an angle
of 60° to the horizontal. The force of friction
on the crate is 125 N. The mass of the crate is
300 kg. Draw the free body diagram for the
system and calculate the acceleration of the
crate. Calculate the work done by each of the
forces in 'displacing the crate by 5.0 m. Which
of these forces is a "no-work" force ?
ball of mass 0.3 kg moving with a
speed of 6.0 ms-1strikes a wall at an
angle of 60° to the wall. It then
rebounds at the same speed and the
same angle. It is in contact with the
wall for 10 ms. Calculate the impulse
and the average force.
A rock of mass 500 kg slides down from rest
along a hill slope that is 500 m long and
300 m high and reaches the bottom. The
coefficient of kinetic friction, between the
rock and the hill slope is 0.25. Calculate
(i) the potential energy of the rock just
before sliding.
(ii) the work done on the rock by the
frictional force.
(iii) the speed of the rock at the bottom of
the hill,
Draw the free body diagram.
g=10ms2
Answers
Answered by
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ANSWER.......................
The weight acts down and is perpendicular to the movement.
It doesn't do any "work"The Floor acts horizontally, opposing
motionNormally, I would consider the friction to be proportional to the Normal force and do some calculations to figure out the normal force, but here they just tell you that there
is opposition to motion that acts horizontally
, so let's go with that...The man acts at a
angle, but only the portion of the force he
exerts horizontally does any "work" to make
the crate move...so we have to figure out the horizontal part of his force.A triangle
(vector) diagram would be set up
so that the man's force is the hypothenuse, and the horizontal
component is adjacent to the angle
(38 degrees). The vertical component is the opposite leg.If you remember your triangle trigonometry
, this means the the magnitude of the horizontal vector is F*(cos θ)
Our balanced force equation is then:
M*a = F*(cos θ) - f,
where f is the resistance from the floor
(a) a = {F*(cos θ) - f} / M
={450*(0.788)-125}/315
=0.729 meters/s2
(b) if the weight is 315 N, then the mass is 315 = M*(9.81)
solve for M = 32.11kg
and plug this in to the equation
a = {450*(0.788)-125}/32.1 = 7.15 m/s2
The weight acts down and is perpendicular to the movement.
It doesn't do any "work"The Floor acts horizontally, opposing
motionNormally, I would consider the friction to be proportional to the Normal force and do some calculations to figure out the normal force, but here they just tell you that there
is opposition to motion that acts horizontally
, so let's go with that...The man acts at a
angle, but only the portion of the force he
exerts horizontally does any "work" to make
the crate move...so we have to figure out the horizontal part of his force.A triangle
(vector) diagram would be set up
so that the man's force is the hypothenuse, and the horizontal
component is adjacent to the angle
(38 degrees). The vertical component is the opposite leg.If you remember your triangle trigonometry
, this means the the magnitude of the horizontal vector is F*(cos θ)
Our balanced force equation is then:
M*a = F*(cos θ) - f,
where f is the resistance from the floor
(a) a = {F*(cos θ) - f} / M
={450*(0.788)-125}/315
=0.729 meters/s2
(b) if the weight is 315 N, then the mass is 315 = M*(9.81)
solve for M = 32.11kg
and plug this in to the equation
a = {450*(0.788)-125}/32.1 = 7.15 m/s2
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