Math / Calculus

Question52. (II) A $450-\mathrm{kg}$ piano is being unloaded from a truck by rolling it down a ramp inclined at $15^{\circ}$ . 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

STEP 1

1. The piano has a mass of $450 \, \text{kg}$ .
2. The ramp is inclined at an angle of $15^\circ$ .
3. The ramp is $4.0 \, \text{m}$ long.
4. The force exerted by the workers is $1020 \, \text{N}$ parallel to the ramp.
5. Friction is negligible.
6. The piano starts from rest.

STEP 2

1. Calculate the gravitational force component along the ramp.
2. Determine the net force acting on the piano.
3. Use Newton's second law to find the acceleration of the piano.
4. Use kinematic equations to find the final velocity of the piano.

STEP 3

Calculate the gravitational force component along the ramp:
$F_{\text{gravity}} = m \cdot g \cdot \sin(\theta)$
where $m = 450 \, \text{kg}$ , $g = 9.8 \, \text{m/s}^2$ , and $\theta = 15^\circ$ .
$F_{\text{gravity}} = 450 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \sin(15^\circ)$

STEP 4

Determine the net force acting on the piano:
$F_{\text{net}} = F_{\text{gravity}} - F_{\text{workers}}$
where $F_{\text{workers}} = 1020 \, \text{N}$ .

STEP 5

Use Newton's second law to find the acceleration of the piano:
$F_{\text{net}} = m \cdot a$
Solve for $a$ :
$a = \frac{F_{\text{net}}}{m}$

STEP 6

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

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