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The displacement x of a body of mass 1 k...

The displacement x of a body of mass 1 kg on a horizontal smooth surface as a function of time t is given by `x = (t^4)/4`. The work done in the first second is

A

`1/4 J`

B

`1/2 J`

C

`3/4 J`

D

`5/4 J`

Text Solution

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The correct Answer is:
To find the work done on a body of mass 1 kg during the first second, given the displacement function \( x(t) = \frac{t^4}{4} \), we can follow these steps: ### Step 1: Find the velocity as a function of time The velocity \( v(t) \) is the derivative of the displacement \( x(t) \) with respect to time \( t \). \[ v(t) = \frac{dx}{dt} = \frac{d}{dt} \left( \frac{t^4}{4} \right) = t^3 \] ### Step 2: Calculate the initial and final velocities We need to find the velocities at \( t = 0 \) seconds and \( t = 1 \) second. - At \( t = 0 \): \[ v(0) = 0^3 = 0 \, \text{m/s} \] - At \( t = 1 \): \[ v(1) = 1^3 = 1 \, \text{m/s} \] ### Step 3: Calculate the change in kinetic energy The kinetic energy \( KE \) of an object is given by the formula: \[ KE = \frac{1}{2} m v^2 \] We will calculate the kinetic energy at \( t = 0 \) and \( t = 1 \): - At \( t = 0 \): \[ KE(0) = \frac{1}{2} \times 1 \times (0)^2 = 0 \, \text{J} \] - At \( t = 1 \): \[ KE(1) = \frac{1}{2} \times 1 \times (1)^2 = \frac{1}{2} \, \text{J} \] ### Step 4: Calculate the work done According to the work-energy theorem, the work done \( W \) is equal to the change in kinetic energy: \[ W = KE(1) - KE(0) = \frac{1}{2} \, \text{J} - 0 \, \text{J} = \frac{1}{2} \, \text{J} \] Thus, the work done in the first second is \( \frac{1}{2} \, \text{J} \). ### Final Answer The work done in the first second is \( \frac{1}{2} \, \text{J} \). ---
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