Question

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.

Answer 1

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.

Answer 2

CORRECTION:

5. Wn = Fn * d = 32.49 * 24.6 = 799J.

Answer 3

friction cant be positive

Answer 4

i have the same numbers and tried ur answer for number 2 and its wrong.... so it would be -26.131*24.6?