Work
Work – force acting on an object – transfer of energy
1) net force exerted on object
2) object is displaced by force
3) net force is same as displacement
W = F ∆d
Unit = 1 N • m = 1 J (scalar)
Example 1
A mother pushes a stroller 42.5 m with a horizontal force of 123N.
Calculate how much work is done.
Given:
∆d = 42.5 m
F = 123N
Required:
W=?
Analysis:
W = F ∆d
Substitute:
Solve:
= (123 N) (42.5 m)
= 5227.5 J = 5.23 x 103 J
Calculating Work with ANGLED Forces:
A boy tries to move a box by pushing down and right.
Box moves horizontally so only X component counts.
If a Force is applied & NOTHING happens – NO WORK is done!
Positive work – Force is in the SAME direction as object motion
W = F ∆d
Negative Work – Force is in OPPOSITE direction as object
motion
W = - F ∆d
Work – difference in energy between two reference points
W = E2 – E1
Example 1
A student is pulling a toboggan at a constant velocity with a force of
87.0 N. The rope handle of the toboggan is angled at 41° above the
horizontal. How much work is done by the student if he must walk
0.300 km to the toboggan hill?
Given: Fapp = 87 N
θ = 41°
d = 0.30 km = 300 m
Required:
W=?
Analysis:
W = (F cos θ) ∆d
Substitute:
Solve:
= (87 N cos 41°) (300 m)
= (65.66 N) (300 m)
= 19 698 J = 1.97 x 104 J
Example 2
Suppose you carry a box with a weight of 34 N a distance of 4.5 m
across the room. How much work is done?
Given:
∆d = 4.5 m
F = 34N
θ = 90
Required:
W=?
Analysis:
W = F ∆d cos θ
Substitute:
Solve:
= (34 N) (4.5 m) (cos90)
=0J
Example 3
A car is moving along a straight road
when the driver suddenly applies the
brakes. The force of friction between
the tires and the road is 1200 N. The
car comes to a stop in 88 m. Find the
work done by friction.
Given:
∆d = 88 m Ff = 1200N
Required:
W=?
Analysis:
W = Ff ∆d cos θ
Substitute:
Solve:
= (1200 N) (88 m) (cos180)
= -105,600 J
θ = 180
Note: Work is a scalar.
Make F &d Δd positive in the
work formula. USE THE
ANGLE TO DETERMINE
THE sign. A NEGATIVE work
means the force is directed
OPPOSITE to the object’s
MOTION.
Example 4
An ice skater slides to a stop by pushing
her blades against the ice as shown in
Figure 6. The ice exerts a constant force
of 95 N on the skater, and the skater stops
in 1.2 m. The angle between the force and
the skater’s direction of motion is 140°.
Find the work done on the skater by the ice.
Given:
∆d = 1.2 m Ff = 95N
θ = 140
Required:
W=?
Analysis:
W = Ff ∆d cos θ
Substitute:
Solve:
= (95 N) (1.2 m) (cos140)
= -87 J
Example 5
A stone on the end of a rope is whirled in a
horizontal circle. The centripetal force is 50
N. How much work is done as the stone
covers a distance of 2 m?
Given:
∆d = 2 m
FC = 50N
θ = 90
Required:
W=?
Analysis:
W = FC ∆d cos θ
Substitute:
Solve:
= (50 N) (2 m) (cos90)
=0J
Note: No work is
done when an
object moves in
a circle
Example 6
The hiker exerts a constant force of 135 N
on the sled at a 48° angle to the sled’s
displacement. The force of friction is
67.0 N. Calculate the work done by the
hiker, the work done by friction, and the
total work done on the sled when the hiker
pulls the sled 345 m over the snow.
Given:
∆d = 345 m Fapp = 135N θ = 48
Required:
W=?
Analysis:
Wapp = Fapp ∆d cos θ
Substitute:
Analysis:
Substitute:
Analysis:
Substitute:
Ff = 67.0N
Solve:
= (135 N) (345 m) (cos48) = 31164.76 J
Wf = Ff ∆d cos θ
Solve:
= (67.0 N) (345 m) (cos180) = -23115 J
WT = Wapp + Wf
Solve:
= 31164.76 J + -23115 J = 8049.76 J = 8.05 x 103 J
Example 7
From what height must a 10 kg hammer fall in
order to do 240 J of work on a stake being driven
into the ground? (Hint: the gravitational force must
first do 240 J of work on the hammer.)
Given:
W = 240 J Fg = mg
θ = 90 g
= 9.81 m/s2
Required:
∆d = ?
Analysis:
W = Fg ∆d cos θ = mg ∆d cos θ
∆d = W / mg cos θ
Substitute:
Solve:
= 240 J / (10kg)(9.81 m/s2)(cos0)
= 2.4 m