A student standing on a scale in an elevator at rest sees that his weight is 840 N. As the elevator rises his weight increases to 1050 N, then returns to normal. When the elevator slows to a stop at the 10th floor his weight drops to 588 N then returns to normal. Determine the acceleration at the beginning and end of the trip.
2006-12-06 10:30:11 · 2 answers · asked by w_xsoadx_w 2 in Science & Mathematics ➔ Physics
F=m*a
When the elevator is at rest the weight is 840N
so
840=m*g
840/9.81=m
the acceleration at the beginning will be
1050 = 840 +a*840/9.81
((1050/840)-1)*9.81=a
at the end
588=840-a*840/9.81
(1-588/840)*9.81=a
j
2006-12-06 10:53:31 · answer #1 · answered by odu83 7 · 0⤊ 0⤋
F = mass * (sum of the acceleration)
840 N = mass*g, the mass is 840N/g
1050 N = 840 N + (840/g)a, solve for a
588 N = 840 N + (840/g)a, solve for a
sorry no calculator
2006-12-06 20:29:13 · answer #2 · answered by Anonymous · 0⤊ 0⤋