Asked by Iki
Maricruz slides a 66 kg crate, m, up an inclined ramp 2.0 m long and attached to a platform 1.0 m above floor level, as shown in Figure 10-19. A 415 N force, F, parallel to the ramp, is needed to slide the crate up the ramp at a constant speed.
(a) How much work does Maricruz do in sliding the crate up the ramp?
(b) How much work would be done if Maricruz simply lifted the crate straight up from the floor to the platform?
(a) How much work does Maricruz do in sliding the crate up the ramp?
(b) How much work would be done if Maricruz simply lifted the crate straight up from the floor to the platform?
Answers
Answered by
Henry
Fp = 415N. = Force parallel to the ramp
a. W = Fp*d = 415 * 2 = 830 Joules.
b. Wc = mg = 66kg * 9.8N/ikg = 647N. =
Weight of crate.
Work = Fc*h = 647 * 1 = 647 Joules.
a. W = Fp*d = 415 * 2 = 830 Joules.
b. Wc = mg = 66kg * 9.8N/ikg = 647N. =
Weight of crate.
Work = Fc*h = 647 * 1 = 647 Joules.
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