SME
Grade 11 Science · Problem type
Work done by a pull at an angle
Use this for “A crate is pulled 15 m across a floor by a 40 N force at 30° above level. How much work does the force do?”
Diagram
Weight = m × g = 5 × 9.8 = 49 N
Normal force = Weight − F sin 30° = 49 − 40 × 0.5 = 29 N
Net force = F cos 30° − f = 34.6 − 0 = 34.6 N
a = Net force/m = 34.6/5 = 6.93 m/s²
Work done by the pull: W = Fd cos θ = 40 × 15 × cos 30° = 520 J
Only the part of the pull along the motion does work; the weight and the normal force are at right angles to it and do none.
This is an example. Type your own values to start a new problem.
W (J)
=
F (N)
×
d (m)
×
cos(
θ
°)
m
Mass of the crate
example
Fg
Weight
calculated
FN
Normal force
calculated
Formulas
Fg
=
m
×
9.8
49 = 5 × 9.8
FN
=
Fg
−
F
×
sin(
θ
)
29 = 49 − 40 × sin(30°)
W
=
F
×
d
×
cos(
θ
)
520 = 40 × 15 × cos(30°)
Assumptions
•
Only the part of the force along the motion, F cos θ, does work: W = Fd cos θ.
•
At θ = 0 the whole force is along the motion; at 90° it does no work.
•
The weight and the normal force are at right angles to the motion, so they do no work.
Step-by-step
WE KNOW
F = 40 N, θ = 30°, d = 15 m, m = 5 kg
FIND
weight (Fg), normal force (FN), work done by the pull (W)
Step 1 · Find weight (Fg)
Use Fg = m × 9.8
The weight is the mass times g.
Fg = 5 × 9.8
Fg = 49 N
Step 2 · Find normal force (FN)
Use FN = Fg − F × sin(θ)
The
pull’s
part
up,
F
sin
θ
,
lifts
a
little:
the
floor
pushes
up
less.
FN = 49 − 40 × sin(30°)
FN = 49 − 40 × 0.5
FN = 49 − 20
FN = 29 N
Step 3 · Find work done by the pull (W)
Use W = F × d × cos(θ)
The
pull’s
part
along
the
floor,
F
cos
θ
,
times
the
distance.
W
=
40
×
15
×
cos(30°)
W
=
40
×
15
×
0.866
W
=
600
×
0.866
W
=
520
J
Check
Put the numbers back in. Both sides should match.
49 = 5 × 9.8 ✓
29 = 49 − 40 × sin(30°) ✓
520 ≈ 40 × 15 × cos(30°) ✓
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