# Problems

A box of mass 5.0 kg is pulled vertically upwards by a force of 68 N applied to a rope attached to the box. Find a) the acceleration of the box and b) the vertical velocity of the box after 2 seconds.

Solution:

a)
From the 2nd Law:
 ma = T - mg a = - g = - 9.8 m/s 2 = 3.8 m/s 2 (5)

b)
Since a is constant
 v = v0 + at = 0 + 3.8(2) = 7.6  m/s. (6)

A hockey puck of mass .5 kg travelling at 10 m/s slows to 2.0 m/s over a distance of 80 m. Find a) the frictional force acting on the puck and b) the coefficient of kinetic friction between the puck and the surface.

Solution:

a)
First we find the acceleration of the puck from the kinematic equations of motion. We have, v0 = 10  m/s , v = 2  m/s and x = 80  m . The third equation of motion gives,
 v 2 = v02 + 2ax a = = = - 0.6  m/s 2 (7)
From the Second Law:
In the x-direction,
 fk = ma = .5(- 0.6) = - 0.3  N. (8)
b)
Use fk = - N . From the y-component of the 2nd Law: N - mg = 0 . Combining,
 - mg = ma = - a/g = 0.061. (9)

A student of mass 50 kg tests Newton's laws by standing on a bathroom scale in an elevator. Assume that the scale reads in newtons. Find the scale reading when the elevator is a) accelerating upward at .5 m/s 2, b) going up at a constant speed of 3.0 m/s and c) going up but decelerating at 1.0 m/s 2.

Solution:

From the 2nd Law:

 Fs - mg = ma Fs = m(g + a). (10)
This gives:
a)
Fs = 50(9.8 + 0.5) = 515  N
b)
Fs = 50(9.8 + 0) = 490  N
c)
Fs = 50(9.8 - 1.0) = 440  N

A wooden plank is raised at one end to an angle of 30 o . A 2.0 kg box is placed on the incline 1.0 m from the lower end and given a slight tap to overcome static friction. The coefficient of kinetic friction between the box and the plank is = 0.20 . Find a) the rate of acceleration of the box and b) the speed of the box at the bottom. Assume that the initial speed of the box is zero.

Solution:

a)
Find the components of the weight of the object:
 wx = - mgsin wy = - mgcos .
Write out the two components of Newton's 2nd Law:
 x : - mgsin + fk = - ma y : N - mgcos = 0. (11)
Using fk = N we get,
 ma = - (mgcos ) + mgsin a = g(sin - cos ) = 9.8(sin 30 - 0.2cos 30) = 3.20 m/s 2 (12)
b)
Since a is constant and v 2 = v02 + 2ax . With x = 1  m , v0 = 0 we have
 v = = 2.53  m/s .

A 10 kg box is attached to a 7 kg box which rests on a 30 o incline. The coefficient of kinetic friction between each box and the surface is = .1 . Find a) the rate of acceleration of the system and b) the tension in the rope.

Solution:

We apply the 2nd law separately to each box.
For the 10 kg box:
y direction:

 N2 - m2g = N2 = m2g,
x direction:
 T - fk 2 = m2a T - N2 = m2a T - m2g = m2a. (13)
For the 7 kg box:
y direction:
 N1 = m1gcos ,
x direction:
 m1gsin - T - fk 1 = m1a m1gsin - T - N1 = m1a m1gsin - T - m1gcos = m1a. (14)

We have a system of two equations and two unknowns: a and T . We can solve as follows.

a)
From equation (4.1), T = m2a + m2g . Substituting into equation (4.2) gives,
 m1a = m1gsin - m1gcos - m2a - m2g m1a + m2a = m1gsin - m2g - m1gcos (15)

 a = [m1gsin - m2g - m1gcos ]g] = [7(9.8)sin 30 - (0.1)(10)(9.8) - (0.1)(7)(9.8)cos 30)] = 1.1 m/s 2 (16)
b)
Then substituting into the first equation gives,
 T = m2(a + g) = 10(9.8(.1) + 1.1) = 20.8  N. (17)