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lesantik [10]
4 years ago
6

A crate is pulled across a horizontal, frictionless floor by a rope. At the same time, the crate pulls back on the rope, in acco

rd with Newton's third law. Does the work done on the crate by the rope then equal zero?
Physics
1 answer:
Verdich [7]4 years ago
4 0

Answer:

Explanation:

If force F is applied on the crate , it will move with acceleration

a = F / m as no other force is acting on it .

This force is applied on the crate . work done by rope on crate  will be positive and equal to F x d where d is displacement .

The reaction force will apply on hand through rope. So work will also be done on hand by crate . This work will be negative as displacement in hand is opposite to direction of force on it.

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7 0
3 years ago
What are Sir Issac Newton's three laws of motion?
Likurg_2 [28]
The first law is that every object stay at rest or stay in uniform motion in a straight line until it is forced to change its state by the action of an external force. This law is called law of inertia.

The second law is that the acceleration of an object is dependent upon two variables. the net force acting upon the object and the mass of the object.  F= ma or force is equal to mass times acceleration. This law is known as the law of force and acceleration.  

The third law is that for every action there is an equal and opposite reaction.  every interaction there is a pair of forces acting on the two interacting objects. the size of forces on the first object equals the size of the force on the second object. 

Hope this helps :) 

can you please make this the brainliest answer it would really help . Thanks
4 0
3 years ago
Read 2 more answers
A solid sphere of weight 42.0 N rolls up an incline at an angle of 36.0°. At the bottom of the incline the center of mass of the
Alecsey [184]

Answer:

Part a)

KE = 77.95 J

Part b)

L = 3.16 m

Part c)

distance L is independent of the mass of the sphere

Explanation:

Part a)

As we know that rotational kinetic energy of the sphere is given as

KE = \frac{1}{2}I\omega_2 + \frac{1}{2}mv^2

so we will have

KE = \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 + \frac{1}{2}mv^2

so we will have

KE = \frac{1}{5} mv^2 + \frac{1}{2}mv^2

KE = \frac{7}{10} mv^2

KE = \frac{7}{10}(\frac{42}{9.81})(5.10^2)

KE = 77.95 J

Part b)

By mechanical energy conservation law we know that

Work done against gravity = initial kinetic energy of the sphere

So we will have

mgLsin\theta = KE

\frac{42}{9.81}(9.81)L sin36 = 77.95

L = 3.16 m

Part c)

by equation of energy conservation we know that

\frac{7}{10}mv^2 = mgL sin\theta

so here we can see that distance L is independent of the mass of the sphere

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evablogger [386]

Answer:

Explanation:

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