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sergij07 [2.7K]
3 years ago
11

A roller coaster begins at the top of a hillIf it accelerates at the rate of 2m/s ^2 And has a mass of 2000kg, what net force is

acting on it? (F=ma)
Physics
1 answer:
Luden [163]3 years ago
5 0

The correct answer to the question is : 4000 N

EXPLANATION:

As per the question, the mass of the roller coaster is given as m = 2000 kg

The acceleration of the roller coaster is given as a = 2m/s2

We are asked to calculate the net force acting on the roller coaster.

From Newton's second laws of motion,we know that net external force acting on a particle is equal to the product of mass with acceleration of the particle.

Mathematically it is written as-

F = ma

Hence,net force acting on the roller coaster is calculated as-

F = 2000Kg×2m/s2

= 4000 N.

Here, Newton (N) is the unit of force.

Hence,net force acting on the roller coaster is 4000 N.

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A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.
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Answer: f_{r} = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

W_{x} = horizontal component of the weight = mgsinФ

W_{y} = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - f_{r} = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8m/s^{2}

f_{r} = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

v^{2} = u^{2} + 2aS

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}

we slot in a into the equation below to get frictional force

mgsinФ - f_{r} = ma

3 * 9.8 * sin 37 - f_{r} = 3* 0.4

17.9633 - f_{r} =  1.2

f_{r} = 17.9633 - 1.2

f_{r} = 16.49N

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