The man pushes with a force of 200 N. The child and sled
combo have a mass of 30 kg and the coefficient of kinetic friction is
0.15.The angle between the slide and the ground is 30 degrees.What is the frictional force opposing his efforts?
What is the acceleration of the child?
To find the frictional force opposing his efforts, we use the formula:
frictional force = kinetic friction coefficient * normal force
First, we need to find the normal force. The weight of the child and sled is:
weight = mass * gravitational acceleration
weight = 30 kg * 9.8 m/s^2
weight = 294 N
The normal force is equal in magnitude and opposite in direction to the weight, so:
normal force = -294 N
(Note that we use a negative sign to indicate that the normal force is in the opposite direction to the weight, i.e. perpendicular to the slope.)
The frictional force is therefore:
frictional force = 0.15 * (-294 N)
frictional force = -44.1 N
(Note again that we use a negative sign to indicate that the frictional force is in the opposite direction to the applied force, i.e. parallel to the slope.)
To find the acceleration of the child, we use Newton's second law:
force net = mass * acceleration
The net force acting on the child is the horizontal component of the applied force minus the frictional force:
force net = force applied * cos(angle) - frictional force
force net = 200 N * cos(30 degrees) - (-44.1 N)
force net = 200 N * 0.866 - (-44.1 N)
force net = 207.2 N
(Note that we use the cosine of the angle between the slide and the horizontal to find the horizontal component of the applied force.)
The acceleration of the child is therefore:
acceleration = force net / mass
acceleration = 207.2 N / 30 kg
acceleration = 6.91 m/s^2
So the frictional force opposing his efforts is -44.1 N and the acceleration of the child is 6.91 m/s^2.