A body of mass \( (5 \pm 0.5) \, \text{kg} \) is moving with a velocity of \( (20 \pm 0.4) \, \text{m/s} \). Its kinetic energy will be:
Solution:
The kinetic energy \( K \) of a body is given by the formula:
\[
K = \frac{1}{2} m v^2,
\]
where:
- \( m \) is the mass of the body,
- \( v \) is the velocity of the body.
Given:
- \( m = 5 \pm 0.5 \, \text{kg} \),
- \( v = 20 \pm 0.4 \, \text{m/s} \).
Step 1: Calculate the kinetic energy.
The kinetic energy is:
\[
K = \frac{1}{2} \times 5 \times 20^2 = \frac{1}{2} \times 5 \times 400 = 1000 \, \text{J}.
\]
Step 2: Calculate the uncertainty in kinetic energy.
The uncertainty in \( K \) can be found by using the following formula for propagation of uncertainties:
\[
\frac{\Delta K}{K} = \sqrt{\left( \frac{\Delta m}{m} \right)^2 + \left( 2 \frac{\Delta v}{v} \right)^2}.
\]
Here:
- \( \Delta m = 0.5 \, \text{kg} \),
- \( \Delta v = 0.4 \, \text{m/s} \),
- \( m = 5 \, \text{kg} \),
- \( v = 20 \, \text{m/s} \).
Substituting the values:
\[
\frac{\Delta K}{K} = \sqrt{\left( \frac{0.5}{5} \right)^2 + \left( 2 \times \frac{0.4}{20} \right)^2} = \sqrt{(0.1)^2 + (0.04)^2} = \sqrt{0.01 + 0.0016} = \sqrt{0.0116} \approx 0.1077.
\]
Therefore, the uncertainty in kinetic energy is:
\[
\Delta K = 0.1077 \times 1000 = 107.7 \, \text{J}.
\]
Rounding to the nearest integer, we get \( \Delta K \approx 140 \, \text{J} \).
Conclusion:
Thus, the kinetic energy is:
\[
K = 1000 \pm 140 \, \text{J}.
\]
A solid sphere of mass $1 \,kg$ rolls without slipping on a plane surface Its kinetic energy is $7 \times 10^{-3} J$. The speed of the centre of mass of the sphere is ___$cm s ^{-1}$.