A 15.4kg block is dragged over a rough horizontal surface by a constant force 191N acting at an angle of 25.3 degrees above the horizontal. The block is displaced 80.4m and the coefficient of kinetic friction is. 2
A) find the magnitude of the work done by the force of friction.
B)what is the work done on the block
In joules
Frg
To find the magnitude of the work done by the force of friction and the work done on the block, we first need to calculate the net force acting on the block.
A) To find the net force, we need to break down the given force into its horizontal and vertical components:
Horizontal component of the force = force * cos(angle)
F_horizontal = 191N * cos(25.3°) = 172.89 N
Vertical component of the force = force * sin(angle)
F_vertical = 191N * sin(25.3°) = 81.28 N
The only other force acting horizontally is the force of friction. The frictional force is given by the equation:
Frictional force = coefficient of kinetic friction * normal force
Since the block is on a horizontal surface, the normal force is equal to the gravitational force acting on the block:
Normal force = mass * gravity
Normal force = 15.4 kg * 9.8 m/s^2 = 150.92 N
Frictional force = 0.2 * 150.92 N = 30.18 N
The net force acting on the block horizontally is:
Net force (horizontal) = F_horizontal - Frictional force
Net force (horizontal) = 172.89 N - 30.18 N = 142.71 N
B) The work done on the block can be calculated using the equation:
Work = force * displacement * cos(angle)
Since the force and displacement are both acting horizontally, and the angle between them is 0°, the equation simplifies to:
Work = Net force (horizontal) * displacement
Work = 142.71 N * 80.4 m = 11,470.84 J
Therefore,
A) The magnitude of the work done by the force of friction is 30.18 J.
B) The work done on the block is 11,470.84 J.