1 kg ball rolls off a 33 m high cliff, and lands 23 m from the base of the cliff. Express the displacement and the gravitational
1 answer:
Answer:
d = <23, 33, 0> m , F_W = <0, -9.8, 0> , W = -323.4 J
Explanation:
We can solve this exercise using projectile launch ratios, for the x-axis the displacement is
x = vox t
Y Axis
y = t - ½ g t²
It's displacement is
d = x i ^ + y j ^ + z k ^
Substituting
d = (23 i ^ + 33 j ^ + 0) m
Using your notation
d = <23, 33, 0> m
The force of gravity is the weight of the body
W = m g
W = 1 9.8 = 9.8 N
In vector notation, in general the upward direction is positive
W = (0 i ^ - 9.8 j ^ + 0K ^) N
W = <0, -9.8, 0>
Work is defined
W = F. dy
W = F dy cos θ
In this case the force of gravity points downwards and the displacement points upwards, so the angle between the two is 180º
Cos 180 = -1
W = -F y
W = - 9.8 (33-0)
W = -323.4 J
You might be interested in
Answer:
17304 J
Explanation:
Complete statement of the question is :
In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 84 kg , are pulled at a constant speed by a tow rope that maintains a constant tension of 350 N .
Part A
How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?
Solution :
= tension force in the tow rope = 350 N
= length of the incline surface = 120 m
= work done by tension force = ?
The tension force acts parallel to incline surface, hence work done by tension force is given as
= height gained by the rider = 30 m
= total mass of rider and tube = 84 kg
Potential energy gained is given as
= Thermal energy created
Using conservation of energy