Question

A 2.0 kg block is pushed 1.0 m at a constant
velocity up a vertical wall by a constant force
applied at an angle of 27.0
◦ with the horizontal, as shown in the figure.
The acceleration of gravity is 9.81 m/s
2.
If the coefficient of kinetic friction between
the block and the wall is 0.40, find
a) the work done by the force on the block.
Answer in units of J.

b) the work done by gravity on the block.
Answer in units of J.

c) the magnitude of the normal force between
the block and the wall.
Answer in units of N.

Answer 1

1) Work done by the force: 8.3 J

2) Work done by gravity: -19.6 J

3) Normal force: 16.3 N

Explanation:

1)

The work done by a force pushing an object is given by

$W=Fd cos \alpha$

where

F is the magnitude of the force

d is the displacement of the object

$\alpha$ is the angle between the direction of the force and of the displacement

First of all, we need to find the magnitude of the force F. Since the block is moving vertically at constant velocity (= zero acceleration), the equation of motion of the block along the vertical direction is:

$F sin \theta - \mu N - mg = 0$ (1)

Where

$\theta=27.0^{\circ}$

$\mu N$ is the force of friction, where

$\mu=0.40$ is the coefficient of friction

N is the normal reaction of the wall

(mg) is the weight of the block, where

m =2.0 kg is the mass of the block

$g=9.81 m/s^2$ is the acceleration of gravity

Along the horizontal direction, the equation of motion is:

$F cos \theta = N$ (2)

Substituting (2) into (1),

$F sin \theta - \mu F cos \theta - mg =0$

And solving for F,

$F(sin \theta - \mu cos \theta) = mg\\F=\frac{mg}{sin \theta - \mu cos \theta}=\frac{(2.0)(9.81)}{sin 27^{\circ} - 0.40 cos 27^{\circ}}=18.3 N$

Now we can calculate the work done by the force. Here we have

d = 1.0 m is the displacement

$\alpha = 90^{\circ} - 27^{\circ} = 63^{\circ}$ is the angle between the direction of the force and the displacement

Substituting,

$W=(18.3)(1.0)(cos 63^{\circ})=8.3 J$

2)

The work done by gravity is equal to:

$W_g = mg d cos \alpha$

where

m = 2.0 kg is the mass of the block

$g=9.81 m/s^2$

d = 1.0 m is the displacement

$\alpha=180^{\circ}$, since the direction of gravity is downward while the displacement of the block is upward

Substituting,

$W=(2.0)(9.81)(1.0)(cos 180^{\circ})=-19.6 J$

3)

Equation (2) found in part 1) tells that

$N=F cos \theta$

which means that the normal force between the wall and the block is equal to the horizontal component of the pushing force on the block.

Here we have:

F = 18.3 N is the magnitude of the pushing force

$\theta=27.0^{\circ}$ is the angle of the force with the horizontal

Substituting, we find the normal force:

$N=(18.3)(cos 27^{\circ})=16.3 N$

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Answer 2

(a) The work done by frictional force is 7.0 J

(b) The work done by gravity is 8.9 J.

(c) The magnitude of the normal force on the block and the wall is 17.5 N.

The given parameters;

• mass of the block, m = 2 kg
• distance moved by the block, d = 1.0 m
• angle of inclination of the force, θ = 27⁰
• coefficient of friction, μ = 0.4

The normal force on the block is calculated as follows;

$F_n = mgcos\theta$

The frictional force is calculated as;

$F_k = \mu_ k F_n\\\\F_k = \mu_ k mgcos\theta$

The work done by frictional force is calculated as follows;

$W = F_k d = \mu_k mg cos\theta \times d\\\\W = (0.4)(2)(9.8)cos(27) \times 1\\\\W = 7 .0 \ J$

The work done by gravity is calculated as follows;

$W_n = mgsin\theta \times d\\\\W_n = (2) (9.8)sin(27) \times 1\\\\W_n = 8.9 \ J$

The magnitude of the normal force on the block and the wall

$F_n = mgcos\theta\\\\F_n = (2)(9.8) cos(27) \\\\F_n = 17.5 \ N$

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