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4vir4ik [10]
4 years ago
6

A proton is moving at 425 m/s. (a) How much work must be done on it to stop it? (A proton has a mass of 1.67×10−27 kg.) (b) Assu

me the net braking force acting on it has magnitude 8.01×10−16 N and is directed opposite to its initial velocity. Over what distance must the force be applied? Watch your negative signs in this problem.
Physics
1 answer:
Tpy6a [65]4 years ago
8 0

Answer:

a)1.51*10^-22joules b) 1.89*10^-7m

Explanation:

Work done to stop the proton = the kinetic energy of the proton = 1/2 mv^2 = 1/2* 1.67*10^-27* 425*425 = 1.51* 10 ^ -22 joules

b) net force acting to stop the proton = 8.01*10^-16

Work done needed to stop the proton = net force acting opposite the motion * distance

Distance covered = need work done/ net force

Distance = 1.51*10^-22/8.01*10^-16= 1.89*10^-7m

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What is the magnitude of the force required to stretch a 20 cm-long spring, with a spring constant of 100 N/m, to a length of 21
GalinKa [24]

The correct answer to the question is: 1 N.

EXPLANATION:

As per the question, the spring constant or the force constant of the spring is given as k = 100 N/m.

The original length of the spring L = 20 cm.

The stretched length of the spring L'= 21 cm.

Hence, the change in length will be-

                              ∆L = L' - L

                                    = 21 cm - 20 cm

                                    = 1 cm

                                    = 0.01 m

We are asked to calculate the magnitude of force acting on the spring .

From Hooke's  law, we know that the restoring force that acts on the spring is proportional to the distance .

Mathematically it can be written as -

                F = - kx.

Here, k is the force constant.

         x is the change in length due to compression or elongation.

The negative sign is due to the fact that it is opposite to the applied force.


Hence, the applied force on the spring is calculated as -

            F = kx

               = k × ∆L

               = 100 N/ m × 0.01 m

               = 1 N.

Hence, the force acting on the spring is 1 N.


                                                   

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