Question

A milk carton with a mass of 2.00 kg is pulled across a table with a horizontal force of 3.00 N. If the coefficient of friction is 0.110, the magnitude of acceleration of the milk carton is ___m/s2.

Answer 1

Answer on Question #41294 – Physics – Mechanics | Kinematics | Dynamics

A milk carton with a mass of $2.00\,\mathrm{kg}$ is pulled across a table with a horizontal force of $3.00\,\mathrm{N}$. If the coefficient of friction is 0.110, the magnitude of acceleration of the milk carton is ___ m/s².

Solution:

$\mathrm{m} = 2\,\mathrm{kg}$ – mass of the carton;

$\mathrm{F} = 3\,\mathrm{N}$ – horizontal force;

$\mu = 0.110$ – coefficient of friction;

$\mathrm{a}$ – acceleration of the carton;

Newton’s second law for the cart along Y-axis:

y: $\mathrm{N} = \mathrm{mg}$ (1)

Newton’s second law for the cart along X-axis:

x: $\mathrm{F} - \mathrm{F}_{\mathrm{fr}} = \mathrm{ma}$ (2)

Formula for the friction:

$\mathrm{F}_{\mathrm{fr}} = \mathrm{N} \cdot \mu$ (3)

(1) in(3):

$\mathrm{F}_{\mathrm{fr}} = \mathrm{mg} \cdot \mu$ (4)

(4) in(2):

$\mathrm{F} - \mathrm{mg} \mu = \mathrm{ma}$

$$\mathrm{a} = \frac{\mathrm{F} - \mathrm{mg} \mu}{\mathrm{m}} = \frac{\mathrm{F}}{\mathrm{m}} - \mathrm{g} \mu = \frac{3\,\mathrm{N}}{2\,\mathrm{kg}} - 9.8\,\frac{\mathrm{N}}{\mathrm{kg}} \cdot 0.110 = 0.42\,\frac{\mathrm{m}}{\mathrm{s}^2}$$

Answer: magnitude of acceleration of the milk carton is $0.42\,\frac{\mathrm{m}}{\mathrm{s}^2}$.

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