Math  /  Calculus

Question52. (II) A 450kg450-\mathrm{kg} piano is being unloaded from a truck by rolling it down a ramp inclined at 1515^{\circ}. There is negligible friction and the ramp is 4.0 m long. Two workers slow the rate at which the piano moves by pushing with a combined force of 1020 N parallel to the ramp. If the piano starts from rest, how fast is it moving at the bottom?

Studdy Solution
Use kinematic equations to find the final velocity of the piano. Since the piano starts from rest, use:
v2=u2+2as v^2 = u^2 + 2a s
where u=0m/s u = 0 \, \text{m/s} (initial velocity), a a is the acceleration found in Step 3, and s=4.0m s = 4.0 \, \text{m} .
Solve for v v :
v=2as v = \sqrt{2a s}
Now, let's calculate the values:
STEP_1: Fgravity=450kg×9.8m/s2×sin(15)450×9.8×0.25881139.6N F_{\text{gravity}} = 450 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \sin(15^\circ) \approx 450 \times 9.8 \times 0.2588 \approx 1139.6 \, \text{N}
STEP_2: Fnet=1139.6N1020N=119.6N F_{\text{net}} = 1139.6 \, \text{N} - 1020 \, \text{N} = 119.6 \, \text{N}
STEP_3: a=119.6N450kg0.266m/s2 a = \frac{119.6 \, \text{N}}{450 \, \text{kg}} \approx 0.266 \, \text{m/s}^2
STEP_4: v=2×0.266m/s2×4.0m v = \sqrt{2 \times 0.266 \, \text{m/s}^2 \times 4.0 \, \text{m}} v=2.1281.46m/s v = \sqrt{2.128} \approx 1.46 \, \text{m/s}
The final velocity of the piano at the bottom of the ramp is:
1.46m/s \boxed{1.46 \, \text{m/s}}

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