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andrey2020 [161]
3 years ago
9

How does work done on an object affect its kinetic energy

Physics
1 answer:
Alexus [3.1K]3 years ago
5 0

According to the work-energy theorem, the work done on an object by a net force equals the change in kinetic energy of the object. Essentially kinetic energy is the energy used for motion. Interestingly, as work is done on an object, potential energy can be stored in that object. A moving object has kinetic energy because work has been done on it. When work is done energy in one form is transferred to the kinetic energy of the moving object. To stop the object again, the same amount of work would have to be done to bring it back to rest.

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Kipish [7]

Answer:

okay

Explanation:

6 0
3 years ago
Read 2 more answers
A 5 kg block is resting on a ramp inclined at 35 degrees above the horizontal. What is the magnitude of the normal force acting
Mamont248 [21]
I'm pretty sure the answer is b 28n hope helps :)
4 0
3 years ago
A dart gun consists of a horizontal spring with k = 52 Newtons/m that is compressed 43
gulaghasi [49]

Answer:

299 m/s^2

Explanation:

When a spring is compressed, the force exerted by the spring is given by:

F=kx

where

k is the spring constant

x is the compression of the spring

In this problem we have:

k = 52 N/m is the spring constant

x = 43 cm = 0.43 m is the compression

Therefore, the force exerted by the spring on the dart is

F=(52)(0.43)=22.4 N

Now we can apply Newton' second law of motion to calculate the acceleration of the dart:

F=ma

where

F = 22.4 N is the force exerted on the dart by the spring

m = 75 g = 0.075 kg is the mass of the dart

a is its acceleration

Solving for a,

a=\frac{F}{m}=\frac{22.4}{0.075}=299 m/s^2

7 0
3 years ago
A constant 20 N force is applied to a 7 kg box to push it along the ground. How
Elis [28]

Answer:

40 joules

Explanation:

Work Done=Force*Distance

6 0
4 years ago
A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h
Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

                    \tau = I \times \alpha

and,       I = \sum mr^{2}

So,      I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}

                       = 4 kg m^{2}

      \tau_{1} = 4 kg m^{2} \times \alpha_{1}

     \tau_{2} = I_{2} \alpha_{2}

So,      I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}

                     = 1 kg m^{2}

 as \tau_{2} = I_{2} \alpha_{2}

                   = 1 kg m^{2} \times \alpha_{2}        

Hence,     \tau_{1} = \tau_{2}

                  4 \alpha_{1} = \alpha_{2}

            \alpha_{1} = \frac{1}{4} \alpha_{2}

Thus, we can conclude that the new rotation is \frac{1}{4} times that of the first rotation rate.

8 0
3 years ago
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