Holding onto a tow rope moving parallel to a frictionless ski slope, a 61.8 kg skier is pulled up the slope, which is at an angl
e of 6.8° with the horizontal. What is the magnitude Frope of the force on the skier from the rope when (a) the magnitude v of the skier's velocity is constant at 2.47 m/s and (b) v = 2.47 m/s as v increases at a rate of 0.109 m/s2?
1 answer:
Answer:
a) Frope= 71.7 N
b) Frope=6.7 N
Explanation:
In the figure the skier is simulated as an object, "a box".
a) At constant velocity we can say that the object is in equilibrium, so we apply the Newton's first law:
∑F=0
Frope=w*sen6.8°
Frope=71.71N
Take into account that w is the weight that is calculated as mass per gravitiy constant:
w=m*g
b) In this case the system has an acceleration of 0.109m/s2. Then, we apply Newton's second law of motion:
F=m*a
F=61.8Kg*0.109m/s2
Frope=6.73N
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a) 2.75 s
The vertical position of the ball at time t is given by the equation
where
h = 4 m is the initial height of the ball
u = 12 m/s is the initial velocity of the ball (upward)
g = 9.8 m/s^2 is the acceleration of gravity (downward)
We can find the time t at which the ball reaches the ground by substituting y=0 into the equation:
This is a second-order equation. By solving it for t, we find:
t = -0.30 s
t = 2.75 s
The first solution is negative, so we discard it; the second solution, t = 2.75 s, is the one we are looking for.
b) -15.0 m/s (downward)
The final velocity of the ball can be calculated by using the equation:
where
u = 12 m/s is the initial (upward) velocity
g = 9.8 m/s^2 is the acceleration of gravity (downward)
t is the time
By subsisuting t = 2.75 s, we find the velocity of the ball as it reaches the ground:
And the negative sign means the direction is downward.