A person pushes a block of mass M = 6.0 kg with a constant speed of 5.0 m/s straight up along a flat surface inclined 30.0° above the horizontal. The coefficient of kinetic friction between the block and the surface is μ = 0.40. What is the net force acting on the block?
Question
A person pushes a block of mass M = 6.0 kg with a constant speed of 5.0 m/s straight up along a flat surface inclined 30.0° above the horizontal. The coefficient of kinetic friction between the block and the surface is μ = 0.40. What is the net force acting on the block?
Solution
To solve this problem, we need to consider the forces acting on the block: gravity, friction, and the force exerted by the person.
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Calculate the force of gravity: The force of gravity acting on the block is given by F_gravity = m*g, where m is the mass of the block and g is the acceleration due to gravity. So, F_gravity = 6.0 kg * 9.8 m/s^2 = 58.8 N. However, only a component of this force acts along the incline, which is given by F_gravity_incline = F_gravity * sin(30) = 58.8 N * sin(30) = 29.4 N.
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Calculate the force of friction: The force of friction is given by 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 less than the weight of the object, and is given by F_normal = mgcos(30) = 6.0 kg * 9.8 m/s^2 * cos(30) = 50.9 N. So, F_friction = 0.40 * 50.9 N = 20.36 N.
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Calculate the force exerted by the person: Since the block is moving at a constant speed, the net force acting on it is zero. Therefore, the force exerted by the person must balance out the forces of gravity and friction. So, F_person = F_gravity_incline + F_friction = 29.4 N + 20.36 N = 49.76 N.
Therefore, the net force acting on the block is 0 N, because the force exerted by the person balances out the forces of gravity and friction.
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