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A body of 4 kg is lying at rest. under the action of a constant force,it gains a speed of 5m/s.The work done by the force will be Ans-30 joules can you explain it please solve it quicly?
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A body of 4 kg is lying at rest. under the action of a constant force,...
Understanding Work Done by a Force
To determine the work done by a constant force on a body, we can use the work-energy principle. This principle states that the work done on an object is equal to the change in its kinetic energy.
Given Data:
- Mass of the body (m) = 4 kg
- Initial velocity (u) = 0 m/s (since the body is at rest)
- Final velocity (v) = 5 m/s
Step 1: Calculate the Change in Kinetic Energy
The formula for kinetic energy (KE) is:
\[ KE = \frac{1}{2} m v^2 \]
- Initial kinetic energy (KE_initial) when the body is at rest:
\[ KE_{initial} = \frac{1}{2} \times 4 \, \text{kg} \times (0 \, \text{m/s})^2 = 0 \, \text{J} \]
- Final kinetic energy (KE_final) when the body reaches 5 m/s:
\[ KE_{final} = \frac{1}{2} \times 4 \, \text{kg} \times (5 \, \text{m/s})^2 = \frac{1}{2} \times 4 \times 25 = 50 \, \text{J} \]
Step 2: Calculate the Work Done
The work done (W) by the force is the change in kinetic energy:
\[ W = KE_{final} - KE_{initial} \]
Substituting the values:
\[ W = 50 \, \text{J} - 0 \, \text{J} = 50 \, \text{J} \]
Conclusion
The work done by the force is **50 joules**, not 30 joules. If there’s a specific condition or additional force acting that was not mentioned, please clarify.
To determine the work done by a constant force on a body, we can use the work-energy principle. This principle states that the work done on an object is equal to the change in its kinetic energy.
Given Data:
- Mass of the body (m) = 4 kg
- Initial velocity (u) = 0 m/s (since the body is at rest)
- Final velocity (v) = 5 m/s
Step 1: Calculate the Change in Kinetic Energy
The formula for kinetic energy (KE) is:
\[ KE = \frac{1}{2} m v^2 \]
- Initial kinetic energy (KE_initial) when the body is at rest:
\[ KE_{initial} = \frac{1}{2} \times 4 \, \text{kg} \times (0 \, \text{m/s})^2 = 0 \, \text{J} \]
- Final kinetic energy (KE_final) when the body reaches 5 m/s:
\[ KE_{final} = \frac{1}{2} \times 4 \, \text{kg} \times (5 \, \text{m/s})^2 = \frac{1}{2} \times 4 \times 25 = 50 \, \text{J} \]
Step 2: Calculate the Work Done
The work done (W) by the force is the change in kinetic energy:
\[ W = KE_{final} - KE_{initial} \]
Substituting the values:
\[ W = 50 \, \text{J} - 0 \, \text{J} = 50 \, \text{J} \]
Conclusion
The work done by the force is **50 joules**, not 30 joules. If there’s a specific condition or additional force acting that was not mentioned, please clarify.
Community Answer
A body of 4 kg is lying at rest. under the action of a constant force,...
Work energy theorem
W=∆K.E=1/2m(v2-u2). u=0 initially at rest
W=1/2(4kg)(5m/s2)^2=50J
W=∆K.E=1/2m(v2-u2). u=0 initially at rest
W=1/2(4kg)(5m/s2)^2=50J
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