Question

a 450 kg piano is being unloaded from a truck by rolling it down a ramp at 22 degree inclined . there is negligible friction and the ramp is 11.5 m long . two workers slow the rate at which the piano moves by pushing with a combined force of 1420N parralel to the ramp . I f the piano starts fromm rest . how fast is it moving at the bottom???

Answer 1

To solve this problem we will apply the linear motion kinematic equations, and force related equations in Newton's second Law.

Our values are given by:

$m = 450 kg$

$x = 11.5m$

$F = 1420 N$

$\theta = 22\°$

Through equilibrium we know that the Force coming from the vertical weight component minus the Force exerted on the body must be equal to the total Force (Mass and acceleration). So,

$\sum F = ma \\F_w sin(22) - F = ma\\(mg)sin(22) - F = ma\\(450)(9.8)sin(22) - 1420 = 450a\\a = 0.515 m/s^2$

Applying the kinematic equations of motion in which the velocity is described by the change in acceleration and position we have

$v_f^2 = v_i^2 + 2ax$

$v_f^2 = 0 + 2(0.515)(11.5)$

$v_f = 3.44 m/s$

Therefore the Speed at the bottom is 3.44m/s