Gr 12 physics help?

A drop tower ride lifts riders at a constant speed to a height of 78 m and suddenly drops them.
A) determine the work done on a 56 kg rider by the machine as she is lifted to the top of the ride. [Ans. 4.3x10^4 J]
Work = potential energy variation ΔEp = m*g*Δh = 9.8*56*78 = 42,800 joule approx (4.3*10^4 in scientific notation)


B) determine the work done on the rider by gravity as she is lifted to the top[Ans. -4.3x10^4 J]
Such work is exactly the contrary of that done by the machine ...infact :
gravity results in a weight force Fw = m*g directed downward , while lifting machine determines a lifting force F = -m*g , and then directed upward

No reference was made to a lifting angle ...so it is assumed holding straight vertical and sin 90° = 1.00

Newton's II law: F̂ᵣₑsultant = mâ

work done = W = force x distance moved in its direction (line of action)

since the 56 kg person is elevated at constant speed, for dynamic equilibrium (a = 0), the vertical upward force F applied = weight = mg = 56g

=> W = F x 78 = 56 x 9.81 x 78 ~= 4.3 x 10⁴ J

b) W = mg x (-78) ------------------------------------> [-tive cause the distance moved and gravity are oppositely directed]

= -4.3 x 10⁴ J


hope this helps

Here's a hint: F (the force you're working against) is gravity, so F = m*g. Also, if the riders are being lifted directly upward (in the direction of gravity), the angle (theta) is zero.

And yes, that formula is correct for this situation. Just keep in mind the direction of all the forces and distances (i.e., negative signs)