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Calculation Of Force Based On Newton's Second Law Of Motion (2)
Question
1.
The engine
of a vehicle moves it forward with a force of 9600N against a resistive force of
2200N. If the mass of the vehicle is 3400kg, calculate the acceleration
produced.
Solution:
Note: the
net force with which the vehicle accelerated is the difference between the two
forces, i.e.
9600N
– 2200N = 7400 Newtons
Mass = 3400kg, acceleration = ?
Therefore, using the formula for force,
F =
ma
7400 = 3400 x a
a = 7400/3400 = approx. 2.18m/s2
Question
2.
A net force
of 15N acts upon a body of mass 3kg for 5s, calculate the change in the speed of
the body.
Solution:
Given: Force
= 15N, mass = 3kg, time = 5s, ∆v = ?
Using the
force formula, F= m x ∆v/t
Restating
the formula, we have ∆v = Ft/m
∆v =
15 x 5/3 = 25m/s
Question
3.
A bob of a
simple pendulum has a mass of 0.02kg. Determine the weight of the bob. [g = 10ms-2]
Solution:
Given: mass
= 0.02kg, acceleration due to the force of gravity = 10ms-2
Using the
formula for force, F = mg, where g (acceleration due to the force of gravity
replaces a (normal acceleration),
Weight, W = mg
Therefore, W = 0.02 x 10 = 0.2 Newtons
Question
4.
What change
in velocity would be produced on a body of mass 4 kg if a constant force of 16 N
acts on it for 2 s?
Solution:
Given: Force
= 16 Newtons, mass = 4 kg, time = 2 s, change in velocity (∆v) = ?
Using the
formula for force, F = m x ∆v/t
16 = 4
x ∆v/2
Restating
the equation, 16 x 2/4 = ∆v
∆v = 8
m/s
Question
5.
A force of
16 N applied to a 4.0kg block that is at rest on a smooth, horizontal surface.
What is the velocity of the block at t = 5 seconds?
Solution:
Using the
formula for force,
F = m
x ∆v/t or F = m x (v – u)/t
Given: m =
4.0 kg, final velocity v = ?, initial velocity u = 0, time = 5 seconds.
Therefore,
16 =
4.0 x (v – 0)/5
16 =
4.0v/5
4.0v =
80
v =
80/4 = 20
Velocity of the
block is 20 m/s
See calculations of force based on Newton's second law of motion
(3) here.
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