Part 1
A 15.6 kg block is dragged over a rough, hor-
izontal surface by a constant force of 72.2 N
acting at an angle of angle 34.8� above the
horizontal. The block is displaced 24.6 m and
the coefficient of kinetic friction is 0.234.
Find the work done by the 72.2 N force.
The acceleration of gravity is 9.8 m/s2 .
Answer in units of J.
Part 2
Find the magnitude of the work done by the
force of friction.
Answer in units of J.
part 3
What is the sign of the work done by the
frictional force?
1. negative
2. zero
3. positive
part 4
Find the work done by the normal force.
Answer in units of J.
part 5
What is the net work done on the block?
Answer in units of J.
friction cant be positive
CORRECTION:
5. Wn = Fn * d = 32.49 * 24.6 = 799J.
i have the same numbers and tried ur answer for number 2 and its wrong.... so it would be -26.131*24.6?
Part 1:
To find the work done by the 72.2 N force, we can use the formula:
Work = Force * Displacement * cos(theta)
where:
- Force is the magnitude of the force applied, which is 72.2 N
- Displacement is the distance the block is displaced, which is 24.6 m
- theta is the angle between the force and the displacement, which is 34.8 degrees
Using this formula, we can calculate the work done as follows:
Work = 72.2 N * 24.6 m * cos(34.8 degrees)
= 1683.55 J
Therefore, the work done by the 72.2 N force is 1683.55 J.
Part 2:
To find the magnitude of the work done by the force of friction, we can use the formula:
Work = Force of friction * Displacement
where:
- Force of friction is determined by the coefficient of kinetic friction and the normal force. Since we don't have the normal force directly given, we'll need to calculate it first.
- Displacement is 24.6 m (given).
The formula for the force of friction is:
Force of friction = coefficient of kinetic friction * Normal force
where:
- Coefficient of kinetic friction is given as 0.234.
- Normal force can be calculated using the formula:
Normal force = mass * acceleration due to gravity
Substituting the given values:
Normal force = 15.6 kg * 9.8 m/s^2
= 152.88 N
Now we can find the force of friction:
Force of friction = 0.234 * 152.88 N
= 35.762 N
Using this value, we can calculate the magnitude of the work done by the force of friction as follows:
Work = 35.762 N * 24.6 m
= 880.31 J
Therefore, the magnitude of the work done by the force of friction is 880.31 J.
Part 3:
The sign of the work done by the frictional force can be determined by considering the direction of the force and the displacement. In this case, the force of friction is acting in the opposite direction of the displacement (due to the block being dragged), so the work done by the frictional force is negative.
Therefore, the answer is 1. negative.
Part 4:
To find the work done by the normal force, we need to first determine the displacement in the direction of the normal force. However, the normal force acts perpendicular to the displacement, so the work done by the normal force is zero.
Therefore, the work done by the normal force is zero.
Part 5:
The net work done on the block can be obtained by summing up the work done by all the forces acting on it. In this case, we need to consider the work done by the 72.2 N force, the force of friction, and the normal force.
Net Work = Work by 72.2 N force + Work by force of friction + Work by normal force
From the previous calculations:
- Work by 72.2 N force = 1683.55 J
- Work by force of friction = -880.31 J (negative because it opposes the displacement)
- Work by normal force = 0 J
Therefore, the net work done on the block is:
Net Work = 1683.55 J - 880.31 J + 0 J
= 803.24 J
Therefore, the net work done on the block is 803.24 J.
Fb=15.6kg * 9.8N/kg = 152.88N @ 0 deg.=
Force of the block.
Fp = Fh = 155.88sin0 = 0N = Hor comp of
block.
Fv = 155.88cos0 = 155.88N = Force perpendicular to plane.
Applied Force:
Fap = 72.2N @ 34.8deg.
Fh = 72.2cos34.8 = 59.29N. = Hor comp
of applied force.
Fv = 72.2sin34.8 = 41.21N. = Ver comp
of applied force.
Combined Forces:
Ff = u*Fv,
Ff = 0.234(155.88 - 41.21),
Ff = 0.234 * 114.67 = 26.8N. = Force
due to friction.
Fn = Fh - Ff = 59.29 - 26.8 = 32.49 =
Net applied hor force.
1. W = 72.2cos34.8 * d,
W = 59.29 * 24.6 = 1459N.
2. W = Ff * d = 26.8 * 0 = 0. Joules.
3. Positive.
4. W = 41.21 * 0 = 0 Joules.
5. Wn = Fn * d = 26.8 * 24.6 = 659 J.