1. A man pushing a crate of mass m = 92.0 kg at a speed of v = 0.875 m/s encount
ID: 1432635 • Letter: 1
Question
1. A man pushing a crate of mass m = 92.0 kg at a speed of v = 0.875 m/s encounters a rough horizontal surface of length = 0.65 m as in the figure below. If the coefficient of kinetic friction between the crate and rough surface is 0.360 and he exerts a constant horizontal force of 283 N on the crate, find the following.
(a) Find the magnitude and direction of the net force on the crate while it is on the rough surface.
(b) Find the net work done on the crate while it is on the rough surface.
in J
(c) Find the speed of the crate when it reaches the end of the rough surface. in
m/s
2. A 0.33-kg stone is held 1.1 m above the top edge of a water well and then dropped into it. The well has a depth of 4.7 m.
(a) Taking y = 0 at the top edge of the well, what is the gravitational potential energy of the stone–Earth system before the stone is released?
in J
(b) Taking y = 0 at the top edge of the well, what is the gravitational potential energy of the stone–Earth system when it reaches the bottom of the well?
in J
(c) What is the change in gravitational potential energy of the system from release to reaching the bottom of the well?
in J
3. When a 2.00-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches2.63 cm.
(a) What is the force constant of the spring?
in N/m
(b) If the 2.00-kg object is removed, how far will the spring stretch if a 1.00-kg block is hung on it?
in cm
(c) How much work must an external agent do to stretch the same spring 8.70 cm from its unstretched position?
in J
Explanation / Answer
1.solution
(a) By F(net) = F(applied) - F(friction)
=>F(n) = 283 - µ x mg
=>F(n) = 283 - 0.360 x 92 x 9.8
=>F(n) = - 41.576 Newton [-ve indicating that it is a retarding force i.e. opposite the direction of motion]
(b) W = F(n) x s
=>W = 41.576 x 0.65 = 27.024 J
(c) By F = ma
=>41.576 = 92 x a
=>a = 0.45 m/s^2
By v^2 = u^2 - 2as
=>v^2 = (0.875)^2 - 2 x 0.45 x 0.65
=>v = 0.178
=>v = 0.422 m/s
2.solution
2a) Gravitational potential energy is given by
GPE1 = m*g*h1
where h = 1.1 m for the requested gravitational potential energy
GPE1 = 0.33*9.81*1.1
GPE1 = 3.56 J
b) GPE2 = m*g*h2
where h = -4.7 m for the requested gravitational potential energy -- negatuve because it asks for the potential energy when it is lower than the reference which is the top edge of the well.
GPE2 = 0.33*9.81*-4.7
GPE2 = -15.21 J
(c)
change in gravitational potential energy of the system = GPE2 - GPE1
change = -15.21 - 3.56 = -18.56J
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