A 2:35 kg block is pushed 1:57 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of 67.5 degree with the horizontal. The acceleration of gravity is 9.8 m/s2. If the coefficient of kinetic friction between the block and wall is 0.678, and the work done

by F. Answer in units of J.

To find the work done by the force F, we need to calculate the net force acting on the block.

First, we can break down the force F into its horizontal and vertical components. The horizontal component is given by F_h = F * cos(theta), where theta is the angle between the force and the horizontal direction. Plugging in the values, we get F_h = F * cos(67.5 degrees).

Next, we can calculate the gravitational force acting on the block. The weight of the block is given by W = m * g, where m is the mass of the block and g is the acceleration due to gravity. Plugging in the values, we get W = 2.35 kg * 9.8 m/s^2.

Since the block is moving at a constant speed up the wall, the force of gravity and the vertical component of the applied force must cancel out. Therefore, the vertical component of the applied force is equal to the weight of the block, F_v = W.

Now, we can calculate the frictional force acting on the block using the formula F_friction = coefficient of kinetic friction * normal force. The normal force is equal to the weight of the block, so F_friction = coefficient of kinetic friction * W.

The net force acting on the block is the sum of the horizontal component of the applied force and the frictional force, F_net = F_h - F_friction.

Finally, we can calculate the work done by the force F using the formula work = force * distance * cosine(theta), where theta is the angle between the force and the displacement. Since the force is applied horizontally and the displacement is vertical, theta = 90 degrees. Therefore, work = F_net * distance * cos(90 degrees).

Plugging in the values and calculating, we can find the answer in units of Joules (J).

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