The man pulls 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 slides 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 the man's efforts, we first need to calculate the gravitational force acting on the sled and child:
F_gravity = m * g
F_gravity = 30 kg * 9.8 m/s^2
F_gravity = 294 N
Next, we can use the coefficient of kinetic friction and the normal force to calculate the frictional force:
F_friction = friction coefficient * F_normal
F_friction = 0.15 * F_gravity
F_friction = 44.1 N
Therefore, the frictional force opposing the man's efforts is 44.1 N.
To find the acceleration of the child, we need to use Newton's second law:
F_net = m * a
The net force acting on the sled and child is:
F_net = F_applied - F_friction - F_gravity
Substituting the values we know:
F_net = 200 N - 44.1 N - 294 N
F_net = -138.1 N
Since the net force is negative, the sled and child will accelerate in the opposite direction of the force applied. Therefore:
-138.1 N = 30 kg * a
a = -4.6 m/s^2
The acceleration of the child is -4.6 m/s^2 (in the direction opposite to the applied force).