Kinetic energy and force

Question

Where needed in these questions, take the accelaration due to gravity as g=10.0m/s^-2 and the zero of potential energy at the ground level shown.

1. A boy pushes his 20kg sister on a bicycle so that a net force of 20N acts on her. They start from rest.

a) What is her kinetic energy when she is moved 4m?

b) what is her speed?

2.A mass of 2.0kg is moving at a speed of 5.0m/s^-1. A constant force acts on the mass for a distance of 3.0m and reduces the speed to 2.0m/s^-1. Find:

a) the work done by the force

b) the size of the force

c) the maximum distance travelled up the incline

d) the maximum vertical height attained by the mass

3. The figure below shows a path of a stone of mass 0.500kg after being thrown from the point A with a speed of 22.0m/s^-1. Find the value of:

http://www.flickr.com/photos/99973477@N03/9557590011/

a) the total energy of the mass at A just as it leaves that point

b) its gain in potential energy in rising from A to the highest point B

c) its speed at B

d) its total energy at the ground level C, just prior to impact

5. The diagram below represents a track which has been erected to demonstrate certain aspects of the energy of a body. A small cart of mass 4.0kg moves without friction on the track, with a speed of 10m/s^-1 along AB.

http://www.flickr.com/photos/99973477@N03/9560399344/

a) what is the kinetic energy of the cart as it moves along AB?

b) at which point on the track is its kinetic energy a maximum?

c) what is the change in the potential energy of the cart between B and D?

d) what is the kinetic energy of the cart as it moves along the horizontal level DE?

e) calculate the speed of the cart as it passes the point C

Answer 1

1)

Calculate acceleration of the bicycle using second law of the Newton.

            F=ma

           a=F/m

Here F is the net force and m is the mass of the girl

 

Substitute 20 N for F and 20 kg for m in above relation

a=20/20=1 m/s²

Calculate final speed of the sister by using

v²-u²=2as

Here u initial speed and s is distance.

Substitute 0 m/s for u,1 m/s² for a and 4 m for s.

            v²-0²=2*1*4

                 v²=  8

                v=2 √2 m/s

Calculate kinetic energy using

            K.E=(1/2)mv²

Substitute 20 kg for m and 2 √2 m/s for v

K.E=(1/2)*20*(2 √2)²

      =80 J

Answer 2

2)Calculate the acceleration by using

v²-u²=2as

Here u initial speed ,v is the final speed and s is distance.

Substitute 5 m/s for u,2 m/s for v and  3 m for s.

            2²-5²=2*a*3

                 a=  -21/6

                  =-7/2 m/s²

Calculate force by using

            F=ma

Substitute -7/2 m/s²for a and 2 kg for m.

            F=2 kg*(-7/2 m/s²)

             =-7 N

Calculate work done by the force

            W=|F|*s

              =7*3

              =21 J

Work done on incline plane

            W=mgsinθ*h

h is the maximum distance travelled up the incline

Calculate vertical height

h_max=v²/2g

Answer 3

3)

a)

Calculate the total energy of the mass at A

E=(1/2)mv²+mgh

Substitute 0.5 kg for m,22 m/s for v,9.81 m/s² for g and 12 m for h.

            E=(1/2)*0.5*(22)²+0.5*9.81*12

               =121+58.86

               =179.86 J

b)

Calculate the gain potential energy

            PE_gain=mg(h_B-h_A)

Substitute 0.5 kg for m, 9.81 m/s² for g, 12 m for h_A and 16.2 m for h_B

            PE_gain=0.5*9.81*(16.2-12)

                      =20.6 J

c)

Calculate the speed at B by using conservation energy

            PE_A+ KE_A= PE_B+ KE_B

            1/2m(v_A)²+mgh_A=1/2m(v_B)²+mgh_B

           1/2(v_A)²+gh_A=1/2(v_B)²+gh_B

          (v_B)²= (v_A)²+2g(h_A-h_B)

Substitute 22 m/s for v_A, 9.81 m/s² for g, 12 m for h_A and 16.2 m for h_B

            (v_B)²= (22)²+2*9.81*(16.2-12)

                   =566.4

              v_B=23.8 m/s

d)

Calculate velocity at C,          

             (v_C)²= (v_B)²+2g(h_B)                                    

Substitute 23.8 m/s for v_B, 9.81 m/s² for g, and 16.2 m for h_B

               (v_C)²= (23.8)²+2*9.81(16.2)

                       =884.24

              v_C=29.74 m/s

Calculate the total energy of the mass at C

            E_C=(1/2)mv_C²

Substitute 29.74 m/s for v_C, 0.5 kg for m

            E_C=(1/2)*0.5*29.74²

                =221.071 J