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A 7.00-kg object accelerates from rest to a final velocity in 55 seconds. If the magnitude of the

net force acting upon the object is 42.0 N, the final velocity reached by the object is
____________m/s.

User Anderly
by
5.3k points

1 Answer

2 votes

Answer:

The final velocity of the object is 330 m/s.

Step-by-step explanation:

To solve this problem, we first must find the acceleration of the object. We can do this using Newton's Second Law, given by the following equation:

F = ma

If we plug in the values that we are given in the problem, we get:

42 = 7 (a)

To solve for a, we simply divide both sides of the equation by 7.

42/7 = 7a/7

a = 6 m/s^2

Next, we should write out all of the information we have and what we are looking for.

a = 6 m/s^2

v1 = 0 m/s

t = 55 s

v2 = ?

We can use a kinematic equation to solve this problem. We should use:

v2 = v1 + at

If we plug in the values listed above, we should get:

v2 = 0 + (6)(55)

Next, we should solve the problem by performing the multiplication on the right side of the equation.

v2 = 330 m/s

Therefore, the final velocity reached by the object is 330 m/s.

Hope this helps!

User Oleh Dokuka
by
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