Question

A 15.0kg watermelon and a 11.00kg pumpkin are attached each via a cord that wraps over a pulley. Friction is negligible everywhere in the system. Calculate the acceleration of the watermelon and pumpkin. Specify magnitude and direction

Answer 1

Assumption: the angle beneath the pumpkin is 53 degrees and the angle beneath the watermelon is 30 degrees.

Use Newton’s second law to relate the net force to the acceleration.

T – the tension in the cord

Since the pumpkin and the watermelon are attached by the cord, they must have the same magnitude of acceleration.

For the pumpkin:

$m_pgsin53º – T = m_pa$ (1)

For the watermelon:

$T – m_wgsin30º = m_wa$ (2)

Adding (1) and (2) gives

$g(m_psin53º – m_wsin30º) = (m_w + m_p)a$

$a = \frac{g(m_psin53º – m_wsin30º)}{(m_w + m_p)} = \frac{9.8(11\times0.8 – 15\times0.5)}{(15+11)} = 0.49 m/s^2$

The acceleration is positive, so the watermelon slides up the ramp and the pumpkin slides down. The watermelon moves up and to the left.