f 51. If you stand next to a wall on a frictionless skateboand and push the wall
ID: 1884588 • Letter: F
Question
f 51. If you stand next to a wall on a frictionless skateboand and push the wall with a force of 40 N, how hard does the wall push on you? Show that if your mass is 80 kg, your acceleration is 0.5 m/ away from the wall. 52. A force Facts in the forward direction on a cart of mass m. A friction force fopposes this motion. (a) Use Newton's second law and show that the F-1 acceleration of the cart is (b) Show that if the cart's mass is 4.0 kg, the applied force is 12.0 N, and the friction force is 6.0N, the cart's acceleration is 1.5 m/s with an acceleration of 4 m/s.Show that the friction force that acts on the firefighter is 48oN. 54. A rock band's tour bus, mass M, is accelerating away from a stop sign at rate a when a piece of heavy metal, mass M/5, falls onto the top of the bus and remains there. (a) Show that the bus's acceleration is now a (b) If the initial acceleration of the bus is 1.2 m/s, show that when the bus carries the heavy metal with it, the acceleration will be 1.0 m/s 58. Three parachutists, A, B, and C, haveExplanation / Answer
51. From newton's 3rd law,
wall will exert force of same magnitude but in opposite direction.
F = 40 N
a = F/m = 40/80 = 0.5 m/s^2
55. (A) Fnet: A = 5 N to right
B = 5 to right
C = 15- 10 = 5 N to right
D = 15-5 = 10 N to right
D > A = B = C
(b) a = F_net / m
aA = 5/5 = 1 m/s^2
aB = 5/10 = 0.5 m/s^2
aC = 5/5 = 1 m/s^2
aD = 10/20 = 0.5 m/s^2
A = C > B = D
65. F_net = m g = 1 x 9.8 = 9.8 N
now F_net = m g - F_res = 9.8 - 2 = 7.8 N
68. (a) action - normal force by hammer (down)
reaction - normal force by nail on hammer(up)
(b) action - gravitational pull by earth (weight)(down)
reaction- gravitational pull by book on earth (up)
(c) action -> push force by blade on air (downw)
reaction -> push force by air on blade(up)
69 . (a) weight force by earth -> downward
normal reaction by head (upward)
(b) only weight force acting.
81. applying momentum conservation.
pi= pf
0 = m v - 2m v'
v' = v/2
Ans: half the fast as lighter one
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