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
You might be interested in
The text does not specify whether the resistance R of the wire must be kept the same or not: here I assume R must be kept the same.
The relationship between the resistance and the resistivity of a wire is
where
is the resistivity
A is the cross-sectional area
R is the resistance
L is the wire length
the cross-sectional area is given by
where r is the radius of the wire. Substituting in the previous equation ,we find
For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
Therefore, the new resistivity must be 4 times the original one.
The relationship between the resistance and the resistivity of a wire is
where
A is the cross-sectional area
R is the resistance
L is the wire length
the cross-sectional area is given by
where r is the radius of the wire. Substituting in the previous equation ,we find
For the new wire, the length L is kept the same (L'=L) while the radius is doubled (r'=2r), so the new resistivity is
Therefore, the new resistivity must be 4 times the original one.