Question

A 5 kg ball is thrown up with a speed 10m/s (a) Find its kinetic energy at time of thrown. (b) Find its potential energy when it reaches the highest point (c) Height=? (d) What is the work done by the force of gravity?

Answer 1

a)
$ke \: = \: (1 \div 2) \times m \times {v}^{2}$
KE = 250J

b) H = 5m
$pe \: = \: mgh$
PE = 250J

can also be done by COE

c) 10m

d)-250J

Answer 2

Answer:

(a) The kinetic energy, K.E.= 1 2 mv^ 2

= 1 2 *5*100=250]

(b) As the ball rises, its kinetic energy decreases and its potential energy increases. At the topmost point of the motion, the ball comes momentarily at rest. The kinetic energy becomes zero, at this point. The kinetic energy has been converted into potential energy. Therefore P.E at topmost point = K.E. of throw = 250 J.

(c)

P.E.=250J

mgh = 250

h = 250/(5 * 10) = 5m

(d) Work done by the force of gravity, W = Fs * cos theta

W = (mg)(h) * cos theta

W = 5 * 10 * 5 * cos 180 degrees

W = - 250J

(e) The results are:

(i)

Work done by force of gravity = -250 J The change in kinetic energy K.E. final

=

- K.E-initial

=0-250J=-250] .. Work done by force of gravity -Change in kinetic energy

(ii) Work done by force of gravity = -250 J

Change in potential energy - P.E-final - P.E-initial

= 250-0 =250] :. Work done by force of gravity = - Change in potential energy