A driver, traveling at 22.0 m/s, slows down her 2.00 x 103 kg car to stop for a red light. What work is done by the friction for
1 answer:
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Answer:
(a) reaction at each front wheel is 5272N (upward)
(b) force between boulder and pallet is 4124N (compression)
Explanation:
Acceleration of the truck = 1 m/
(to the left)
when the truck moves 1 m to the left, the boulder is B and pallet A are raised 0.5 m, then,
= 0.5 m/
(upward) ,
= 0.5 m/
(upward)
Let T be tension in the cable
pallet and boulder: ∑fy = ∑(fy)eff = 2T- ( +
)g = (
+
)
= 2T- (400 + 50)*(9.81 m/) = (400 + 50)*(0.5 m/
)
T = 2320N
Truck: = ∑(
)eff: =
(3.4m) +
(2.0m) - T (0.6m)=
(1.0m)
Nf = (2.0m)(2000 kg)(9.81 m/ )/3.4m - (0.6 m)(2320 N)/3.4m + (1.0 m)(2000 kg)(1.0 m/
) = 11541.2N - 409.4N - 588.2N = 10544N
∑fy (upward) = ∑(fy)eff: +
-
g = 0
10544 + - (2000kg)(9.81 m/
) = 0
= 9076N
∑fx (to the left) = ∑(fx)eff: - T =
= 2320N + (2000kg)(9.81 m/
) = 4320N
(a) reaction at each front wheel:
1/2 (upward): 1/2 (10544N) = 5272N (upward)
(b) force between boulder and pallet:
∑fy (upward) = ∑(fy)eff: +
g -
= (400kg)(9.81 m/
) + (400kg)(0.5 m/
) = 4124N (compression)