1. A 100N block lies on a frictional surface. A force of 50N was applied horizontally where the block had moved 5 m from the positive direction. Find the work done by the applied force and the work done by the weight of the block.
2. Using Example 1, if the applied force is directed along the positive x direction. Find the work done by the applied force.
Problem Set #2
1. A block of mass 10 kg lying on a horizontal surface is pulled to the right by a force of 10N at an angle of 45° with the horizontal. The coefficient of sliding friction between block and surface is 0.1. Find the total work done on the block when it has moved a distance of 1m from its initial position.
2. Find the Kinetic Energy of a 1000 kg car moving at 25km/hr and when it is moving at 100 km/hr.
1. To find the work done by a force, we use the equation W = F * d * cos(theta), where W is the work done, F is the magnitude of the force, d is the distance displaced in the direction of the force, and theta is the angle between the force and the displacement.
a) The work done by the applied force is calculated by multiplying the magnitude of the force (50N) by the distance moved (5m) and the cosine of the angle between the force and the direction of motion (0°, since the force is applied horizontally). W = 50N * 5m * cos(0°) = 250 Joules.
b) The work done by the weight of the block is calculated by multiplying the magnitude of the weight (100N) by the distance moved (5m) and the cosine of the angle between the weight and the direction of motion (180°, since the weight acts vertically downwards). W = 100N * 5m * cos(180°) = -500 Joules. The negative sign indicates that the work is done against the force of gravity.
2. If the applied force is directed along the positive x direction, there is no angle between the force and the displacement. Therefore, the cosine of the angle is 1. Using the same formula as before, the work done by the applied force is calculated as W = 50N * 5m * cos(0°) = 250 Joules.
Problem Set #2
1. The total work done on the block can be determined by considering the work done by the applied force and the work done against friction.
a) To calculate the work done by the applied force, we need to find the component of the force in the direction of motion. The force of 10N at an angle of 45° can be split into two components: one in the horizontal direction (cosine component) and another in the vertical direction (sine component). The cosine component of the force is calculated as F*cos(theta) = 10N * cos(45°) = 7.07N. Therefore, the work done by the applied force is W = 7.07N * 1m * cos(0°) = 7.07 Joules.
b) The work done against friction can be determined using the equation W = f * d, where f is the force of friction and d is the distance displaced in the direction of motion. The force of friction can be found by multiplying the coefficient of sliding friction (0.1) by the normal force between the block and the surface. The normal force is equal in magnitude and opposite in direction to the weight of the block (mg). Therefore, f = 0.1 * 10kg * 9.8m/s^2 = 9.8N. Hence, the work done against friction is W = 9.8N * 1m = 9.8 Joules.
The total work done on the block is the sum of the work done by the applied force and the work done against friction. Therefore, W_total = 7.07 Joules + 9.8 Joules = 16.87 Joules.
2. The kinetic energy (KE) of an object can be calculated using the formula KE = (1/2) * m * v^2, where m is the mass of the object and v is its velocity.
a) When the car is moving at 25 km/hr, we need to convert the velocity to m/s. Since 1 km = 1000 m and 1 hr = 3600 s, we have 25 km/hr * (1000 m/1 km) * (1 hr/3600 s) = 6.94 m/s. The mass is given as 1000 kg. Using the formula, KE = (1/2) * 1000 kg * (6.94 m/s)^2 = 24,058.7 Joules.
b) When the car is moving at 100 km/hr, we need to convert the velocity to m/s. 100 km/hr * (1000 m/1 km) * (1 hr/3600 s) = 27.78 m/s. The mass is still 1000 kg. Using the formula, KE = (1/2) * 1000 kg * (27.78 m/s)^2 = 385,864.44 Joules.