A 76 kg man starts from rest at the top of a 12.0-m long water slide that is inclined 65° with the horizontal. In sliding down, he encountered a frictional force of 45.0 N. (a) What is the speed of the man at the bottom of the slide? (b) What is his kinetic energy? (c) Suppose the man landed horizontally into the water, how far will he travel before coming to rest? Assume that the frictional force is the same underwater.
a) The speed of the man at the bottom of the slide is 8.2 m/s.
b) The man's kinetic energy is 517 J.
c) The man will travel 4.2 m before coming to rest.
To solve this problem, we will use the principles of Newton's laws of motion and conservation of energy. Let's break down the problem into three parts:
(a) Determine the speed of the man at the bottom of the slide.
(b) Calculate his kinetic energy.
(c) Find out how far he will travel before coming to rest in the water.
(a) To find the speed of the man at the bottom of the slide, we will first calculate the acceleration using Newton's second law of motion:
\[F_{\text{net}} = m \cdot a\]
Where:
\(F_{\text{net}}\) is the net force on the man,
\(m = 76 \, \text{kg}\) is the mass of the man, and
\(a\) is the acceleration.
The net force acting on the man is given by the difference between the gravitational force \(F_g\) and the frictional force \(F_f\):
\[F_{\text{net}} = F_g - F_f\]
Where:
\(F_g = m \cdot g\) is the gravitational force, and
\(g = 9.8 \, \text{m/s}^2\) is the acceleration due to gravity.
The frictional force can be calculated using the coefficient of kinetic friction (\(\mu_k\)) and the normal force (\(F_n\)), where:
\[F_f = \mu_k \cdot F_n\]
The normal force can be determined by decomposing the weight of the man into its components. The normal force will be equal to the component perpendicular to the surface:
\[F_n = m \cdot g \cdot \cos(\theta)\]
Where:
\(\theta = 65^\circ\) is the angle of inclination.
Substituting these values into the equation for the net force:
\[F_{\text{net}} = (m \cdot g) - (\mu_k \cdot (m \cdot g \cdot \cos(\theta)))\]
Now, solve for the acceleration:
\[a = \frac{F_{\text{net}}}{m} = \frac{(m \cdot g) - (\mu_k \cdot (m \cdot g \cdot \cos(\theta)))}{m}\]
Let's calculate the acceleration.
Acceleration:
\[a = \frac{(76 \, \text{kg} \cdot 9.8 \, \text{m/s}^2) - (45 \, \text{N} \cdot (\mu_k) \cdot (76 \, \text{kg} \cdot 9.8 \, \text{m/s}^2) \cdot \cos(65^\circ))}{76 \, \text{kg}}\]