6. A boy pushes a lawn mower (m 17.9 kg) starting from rest across a horizontal
lawn by applying a force of 32.9 N straight along the handle, which is inclined
at an angle of 35.1° above the horizontal. The magnitude of the mower’s accel-
eration is 1.37m/s2, which lasts for 0.58 s, after which the mower moves at a
constant velocity. Determine the magnitude of
(a) the normal force on the mower
(b) the frictional force on the mower
(c) the maximum velocity of the mower
(d) the force applied by the boy needed to maintain the constant velocity
(a) To find the normal force on the mower, we need to consider the forces acting on the mower in the vertical direction. The normal force will equal the weight of the mower:
Weight = mass * gravity
Weight = 17.9 kg * 9.8 m/s^2 = 175.42 N
Since the mower is on a horizontal surface, the normal force will be equal in magnitude but opposite in direction to the weight of the mower. Therefore, the normal force is 175.42 N.
(b) To find the frictional force on the mower, we need to consider the forces acting on the mower in the horizontal direction. The frictional force will be equal in magnitude but opposite in direction to the force applied by the boy. Since the mower is moving at a constant velocity, the frictional force will be equal to the force applied by the boy:
Frictional force = force applied by boy = 32.9 N
(c) To find the maximum velocity of the mower, we need to calculate the work done by the boy in pushing the mower up to that velocity. Since the mower starts from rest and accelerates to a certain velocity, we can use the work-energy theorem:
Work = change in kinetic energy
Work = (1/2) * mass * velocity^2
32.9 N * distance = (1/2) * 17.9 kg * (1.37 m/s)^2
32.9 N * distance = 13.08 J
distance = 13.08 J / 32.9 N
distance = 0.397 m
Velocity^2 = 2 * acceleration * distance
Velocity^2 = 2 * 1.37 m/s^2 * 0.397 m
Velocity = 1.31 m/s
(d) To find the force applied by the boy needed to maintain the constant velocity, we already know that the frictional force is equal to the force applied by the boy. Therefore, the force applied by the boy to maintain the constant velocity is 32.9 N.