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Tcecarenko [31]
3 years ago
15

A 400-kg object has a 100-Newton rightward net force being applied to it. What is the magnitude of the rightward acceleration on

the object (in m/s^2)?
Physics
1 answer:
aliya0001 [1]3 years ago
5 0

Answer:

The answer to your question is a = 0.25 m/s²

Explanation:

Data

mass = m = 400 kg

Force = F = 100 N

acceleration = a = ? m/s²

Process

To solve this problem use Newton's second law that states that the force applied to an object is directly proportional to the mass of the body times its acceleration.

Formula

                       F = ma

solve for a

                       a = \frac{F}{m}

Substitution

                       a = \frac{100}{400}

Simplification and result

                              a = 0.25 m/s²

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Answer:

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Answer:

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A 60kg bicyclist (including the bicycle) is pedaling to the
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a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

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2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force (F_A): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag (R): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

F_{net}=ma

where

F_{net} is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

a=3.1 m/s^2 is its acceleration

Substituting, we find the net force on the bicyclist:

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c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

F_{net}=F_a-R

where:

F_{net} is the net force

F_a is the applied force (forward)

R is the air drag (backward)

In this problem we have:

F_{net}=186 N is the net force (found in part b)

R=60 N is the magnitude of the air drag

Solving for F_a, we find the force produced by the bicyclist while pedaling:

F_a=F_{net}+R=186+60=246 N

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Answer:

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