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The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in Fig. 7.39. At O, the velocity of the ball is 0 m s⁻¹ and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.
Show Answer
Frictionless, so total mechanical energy = value at O =
30 J
(all potential there). At any point,
KE = 30 - PE
and
v
=
2 KE
m
=
4 KE
(since m = 0.5 kg).
At P (PE = 20 J):
KE
= 10
J
⇒ v
=
40
≈ 6.3
m s
-1
.
At Q (PE = 30 J):
KE
= 0
⇒ v
= 0
m s
-1
.
At R (PE = 40 J): this needs 40 J but only 30 J is available, so the ball cannot reach R — it turns back before R. (Read the exact PE values from your copy of Fig. 7.39.)
| Point | Potential energy | Kinetic energy = 30 − U | v = 2K/m , m = 0.5 kg |
|---|---|---|---|
| O | 30 J | 0 J | 0 m s⁻¹ (given) |
| P | 20 J | 10 J |
2 × 10 / 0.5
= 6.3 m s -1 |
| Q | 30 J | 0 J | 0 m s⁻¹ |
| R | 40 J | — | cannot be reached — U > total energy, so the ball never gets to R |
