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hammer [34]
3 years ago
13

A force acts on a body of mass 13 kg initially at rest. The force

Physics
1 answer:
PtichkaEL [24]3 years ago
4 0

Answer:

Force that acted on the body was F = 13 N

Explanation:

If once accelerated, the body covers 60 meters in 6 seconds, then its velocity is 60/6 m/s = 10 m/s

When the force was acting (for 10 seconds) the object accelerated from rest (initial velocity vi = 0) to 10 m/s (its final velocity). therefore we can use the kinematic equation for the velocity in an accelerated motion given by:

v_f=v_i+a*t

which in our case becomes;

10\,m/s=0+a*(10\,s)

and we can solve for the acceleration as:

a = 10/10  m/s^2 = 1 m/s^2

Therefore the force acting on the body, based on Newton's 2nd Law expression: F = m * a is:

F = 13 kg * 1 m/s^2 = 13 N

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An object of mass 4kg is moving along a horizontal plane. If the coefficient of kinetic friction is 0.2 find the friction force
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Answer:

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Explanation:

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3 years ago
A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853
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\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}

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