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Standard XII

Physics

KE & PE Graph

A 1 kg block ...

Question

A 1 kg block at origin has an initial velocity $2 m s^{- I}$ . A force as shown in the F-x graph starts to act on it. If no other forces are acting on it, then find the change in kinetic energy of the block. (Give your answer in SI units)

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Solution

$w = \int f d s =$ Area under $F - X$ curve
$= 3 \times 1 + \frac{1}{2} \times 1 \times 3 - \frac{1}{2} \times 2 \times 3 + \frac{1}{2} \times 1 \times 3 + 3 \times 1$
$= 6 J$
$= w = Δ k$
$= 6 J = Δ k$
$∴$ change in kinetic energy= $6 T$

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A 1 kg block at origin has an initial velocity 2ms−I. A force as shown in the F-x graph starts to act on it. If no other forces are acting on it, then find the change in kinetic energy of the block. (Give your answer in SI units)

w=∫fds= Area under F−X curve=3×1+12×1×3−12×2×3+12×1×3+3×1=6J=w=Δk=6J=Δk∴ change in kinetic energy=6T

A 1 kg block at origin has an initial velocity $2 m s^{- I}$ . A force as shown in the F-x graph starts to act on it. If no other forces are acting on it, then find the change in kinetic energy of the block. (Give your answer in SI units)

$w = \int f d s =$ Area under $F - X$ curve
$= 3 \times 1 + \frac{1}{2} \times 1 \times 3 - \frac{1}{2} \times 2 \times 3 + \frac{1}{2} \times 1 \times 3 + 3 \times 1$
$= 6 J$
$= w = Δ k$
$= 6 J = Δ k$
$∴$ change in kinetic energy= $6 T$