Work Energy and Power Test - 13

W = Fs cos \(\theta\)

⇒ cos \(\theta\) = \(\frac{W}{Fs} = \frac{25}{50}= \frac{1}{2}\)

\(\theta\) = 60°

Mass of 1 body = 5kg

Velocity of 1 body = 10m/s

Mass of 2 body = 20kg

Velocity of 2 body = ?

Velocity of 2 = \(\frac{(mass \;of \;1) \times (velocity \;of \;1)}{mass\; of \;2}\)

 Velocity of 2 = \(\frac{(5) \times (10)}{20}\)

Velocity of 2 = 2.5m/s

Coefficient of restitution is 0.89

As 20% energy is lost during the collision,

\(\frac{mgh'}{mgh}\) = \(\frac{80}{100}\)

\(\frac{h'}{h}\) = 0.8

Coeff of restitution e = \( \sqrt{\frac{h'}{h}}\) = \(\sqrt{0.8}\) = 0.894

Initial momentum = mv

Final momentum = -mv (- sign indicates change in direction)

Net change = initial momentum - final momentum

=mv - (- mv)

= 2mv

(i) Due to the impact or the impulsive force velocities of balls change and hence total K.E. , being a scalar quantity, also changes.

(ii) While momentum being vector is conserved in collision.

The loss of K.E. in impact is responsible for change in temperature.

\(\overline P = \frac{mgh}{t} = \frac{(200)(10)(200)}{3600}\)

= 40000W

= 40 kW

p' = 100% x p + p

= 2p

K' = \(\frac{(p')^2}{2m}\)

= \(\frac{4p^2}{2m}\)

4K = 3K + K

= 300% K + K

KE becomes four times. It means that the kinetic energy increases by 300%.

There is rise in temperature so; mechanical energy is converted into heat energy.

\(\triangle U \) = mgh = 20 x 9.8 x 0.5 

= 98 J

\(acc = \frac{5}{15} = \frac{1}{3}m/s^2\)

\(v = u + at\)

= \(0 + \frac{1}{3} \times 1\)

= \(\frac{1}{3} m/s\)

\(\therefore K.E =\frac{1}{2} \times 15 \times \frac{1}{9}\)

\(= \frac{15}{18} = \frac{5}{6} J\)

Work done = \(\frac{5}{6}\)

We know that, kinetic energy (KE) = \(\frac{1}{2}mv^2\)

Also, force x displacement = \(\frac{1}{2}mv^2\)

F = \(\frac{1}{2}m\frac{v^2}{d} = \frac{KE}{d}\)

The ball rebounds with the same speed. So change in its kinetic energy will be zero i.e., work done by the ball on the wall is zero.

Work done on the wire to strain it will be stored as energy which is converted to heat. Therefore the temperature increases.

(i) The correct answer is "no work done". Here, F is the applied force and s is the displacement.

(ii) The work is to be done when the applied force on the object displaces it to some distance.

(iii) The work done is negative when the body displaces in an opposite direction to the applied force.

Work done by centripetal force is always zero, because force and instantaneous displacement are always perpendicular.

\(W =\vec F. \vec s = F s cos \theta\)

= Fs cos(90°) = 0

Total mass = (50 + 20) = 70 kg

Total height = 20 x 0.25 = 5m\Work done = mgh

= 70 x 9.8 x 5

= 3430 J

Here,

k = \(\frac{F}{x} = \frac{10}{1 \times 10^{-3}}\) = \(10^4 N/m \) 

W = \(\frac{1}{2}kx^2\)

= \(\frac{1}{2} \times 10^4 \times (40 \times 10^{-3})^2\)

= 8J

(i) If two surfaces are coated with lubricant then friction will be reduced so they can slide over each other if one is pushed on the other.

(ii) It is friction which prevents relative motion between two surfaces.

If the surfaces are not moving in relation to each other at the point of contact, because the body is rolling, or if the body is initially stationary and a force is applied that is just barely enough to start it moving, the force of friction is called “static” friction, or “stiction”.