A 6.6 kg box slides down an inclined plane that makes an angle of 40° with the horizontal.If the coefficient of kinetic friction is 0.2, at what rate does the box accelerate down the slope?Express your answer in m/s2, to at least one digit after the decimal point.
Question
A 6.6 kg box slides down an inclined plane that makes an angle of 40° with the horizontal.If the coefficient of kinetic friction is 0.2, at what rate does the box accelerate down the slope?Express your answer in m/s2, to at least one digit after the decimal point.
Solution
To solve this problem, we need to use the following physics equations and concepts:
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The force due to gravity acting on the box is F_gravity = m*g, where m is the mass of the box and g is the acceleration due to gravity. In this case, F_gravity = 6.6 kg * 9.8 m/s^2 = 64.68 N.
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This gravitational force can be resolved into two components: one that acts perpendicular to the inclined plane (F_perpendicular = F_gravity * cos(θ)) and one that acts parallel to the inclined plane (F_parallel = F_gravity * sin(θ)). Here, θ is the angle of the incline. So, F_perpendicular = 64.68 N * cos(40°) = 49.57 N and F_parallel = 64.68 N * sin(40°) = 41.67 N.
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The force of friction acting on the box is F_friction = μ * F_normal, where μ is the coefficient of kinetic friction and F_normal is the normal force. On an inclined plane, the normal force is equal to the component of the gravitational force acting perpendicular to the plane. So, F_friction = 0.2 * 49.57 N = 9.91 N.
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The net force acting on the box is the difference between the component of gravity acting down the slope and the force of friction: F_net = F_parallel - F_friction = 41.67 N - 9.91 N = 31.76 N.
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Finally, we can find the acceleration of the box using Newton's second law, F = m*a. Rearranging for a gives a = F/m = 31.76 N / 6.6 kg = 4.81 m/s^2.
So, the box accelerates down the slope at a rate of 4.81 m/s^2.
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