AT&T; LTE 10:32 AM * 73%- (b) Relative to the configuration with the stone at th

Question

AT&T; LTE 10:32 AM * 73%- (b) Relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stone-Earth system when it reaches the bottom of the well? 1684 (c) What is the change in gravitational potential energy of the system from release to reaching the bottom of the well? Need Help? A 280-N child is in a swing that is attached to a pair of ropes 1.80 m long. Find the gravitational potential energy of the child-Earth system relative to the child's lowest position at the following times. (a) when the ropes are horizontal (b) when the ropes make a 35.0° angle with the vertical (c) when the child is at the bottom of the circular arc A particle moves in the xy plane as shown in the figure below and experiences a friction force with constant magnitude 7.41 N, always in the direction opposite the object's velocity. Calculate the work that you must do to slide the object at constant speed against the friction force as the object moves along the following paths. y (m) 50,5.00) (a) the purple path O to followed by a return purple path to (b) the purple path O to © followed by a return blue path to O (c) the blue path O to© followed by a return blue path to O

Explanation / Answer

Ug = mgh

mass * gravity * height

Since the child's weight is in Newtons, it should already be mass times gravity (one newton is one kgm/s^2)

(A) When the ropes are horizontal, he should be 1.8m above the ground.

Thus:

U = mgh = (280)(1.8) = 504J

(b)

When the ropes make a 35degree angle with the vertical, he should be 1.8 - (1.8cos35) from the ground

= 0.326

thus:

U = 280*0.326 = 91.15 J

(c) Since the child is at the bottom of the arc, unless the chains break, her gravitational potential energy is zero.

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