Question

A student is helping a teacher unload boxes off a truck at a school. A 6.24 kg box is sliding down the ramp at an angle of 36.1°. The ramp exerts a force of kinetic friction of 7.31N on the box. Calculate the acceleration of the box.

Answer 1

Answer on Question #41297, Physics, Mechanics

A student is helping a teacher unload boxes off a truck at a school. A $6.24\mathrm{kg}$ box is sliding down the ramp at an angle of $36.1{}^{\circ}$ . The ramp exerts a force of kinetic friction of 7.31N on the box. Calculate the acceleration of the box.

Solution:

Given:

$m = 6.24\mathrm{kg}$

$\theta = 36.1{}^{\circ}$

$F_{fr} = f = 7.31\mathrm{N},$

$a = ?$

The equation of motion is

$$m a = m g \sin \theta - F _ {f r}$$

where $g$ is the gravity acceleration constant $(9.81\mathrm{m} / \mathrm{s}^2)$ .

The acceleration is

$$a = g \sin \theta - \frac {F _ {f r}}{m}$$

Thus,

$$a = 9. 8 1 \cdot \sin 3 6. 1 {}^ {\circ} - \frac {7 . 3 1}{6 . 2 4} = 4. 6 0 9 = 4. 6 1 m / s ^ {2}$$

Answer. $a = 4.61 \, \text{m/s}^2$ .

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