A 17 kg block is dragged over a rough, hor-

izontal surface by a constant force of 170 N
acting at an angle of angle 32.1◦ above the
horizontal. The block is displaced 33.9 m,
and the coefficient of kinetic friction is 0.199.
a] find the work done by 170N
b] find the magnitude of the work done by the force of friction
c] work done by the normal force
d] what is net work done by the block

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To solve this problem, we need to break it down into smaller steps:

a] To find the work done by the constant force of 170 N, we can start by calculating the component of the force acting in the direction of displacement. The formula to calculate the work done is given by:

Work = Force x Displacement x cos(theta)

Where:
- Work is the work done by the force (in joules, J)
- Force is the magnitude of the force applied (in newtons, N)
- Displacement is the magnitude of the displacement (in meters, m)
- theta is the angle between the force and the displacement (in degrees)

Plugging in the given values:
Force = 170 N
Displacement = 33.9 m
theta = 32.1 degrees

Work = 170 N x 33.9 m x cos(32.1 degrees)
Work = 170 N x 33.9 m x 0.8481
Work ≈ 5144.82 J

Therefore, the work done by the force of 170 N is approximately 5144.82 joules.

b] To find the magnitude of the work done by the force of friction, we use a similar approach. The formula for the work done by the force of friction is:

Work = Force of friction x Displacement

The magnitude of the force of friction can be calculated using the equation:

Force of friction = coefficient of friction x Normal force

Where:
- coefficient of friction is the given coefficient of kinetic friction
- Normal force is the force exerted by the surface perpendicular to the block's motion

To find the normal force, we need to consider that when the block is on a horizontal surface, the weight force of the block is equal in magnitude and opposite in direction to the normal force. Therefore:

Normal force = Weight force = Mass x Gravity

Where:
- Mass is the given mass of the block (17 kg)
- Gravity is the acceleration due to gravity (9.8 m/s²)

Plugging in the given values:
Mass = 17 kg
Gravity = 9.8 m/s²
Coefficient of friction = 0.199
Displacement = 33.9 m

Normal force = 17 kg x 9.8 m/s²
Normal force ≈ 166.6 N

Force of friction = 0.199 x 166.6 N
Force of friction ≈ 33.13 N

Work = 33.13 N x 33.9 m
Work ≈ 1123.07 J

Therefore, the magnitude of the work done by the force of friction is approximately 1123.07 joules.

c] To find the work done by the normal force, we need to consider that the normal force is perpendicular to the displacement. Thus, the work done by the normal force is zero. This is because the dot product of perpendicular vectors is zero.

Therefore, the work done by the normal force is zero joules.

d] The net work done by the block is calculated by subtracting the magnitude of the work done by the force of friction from the work done by the applied force:

Net Work = Work by applied force - Work by friction
Net Work = 5144.82 J - 1123.07 J
Net Work ≈ 4021.75 J

Therefore, the net work done by the block is approximately 4021.75 joules.