Answer F = ma F = 0.80 kg X 1.5 m/s/s = 1.2 N

Answer Given: m = 5.5kg ; a = 4.2 m/s/s, to the right Unknown: Fnet Using Newton's Second Law: Fnet = ma Fnet = (5.5 kg)(4.2 m/s/s) = 23 N, to the right

Answer: The required acceleration: a = ^v/^t (6.7 m/s) / 10 s) = 0.67 m/s/s Then: F = ma m = Fw / g = 61.78 X 10 ^3 N / 9.81 m/s/s = 6.30 X 10^3 kg then: F = ma = (6.30 X 10^3 kg)(0.67 m/s/s) = 4.2 kN Hope You've Been To The Gym !!!

Answer Given: Fa = 75 N ; m = 24 kg Fa = Ff at the point of initial movement Unknown: mu = ? Ff = (mu)(Fn) mu = Ff / Fn therefore: mu = Ff / mg = 75N / 24 kg X 9.81 m/s/s mu = 0.32

Answer Find acceleration first a = F / m = 700 N / 2030 kg = .345 m/s/s Then find the displacement d = vit + 1/2 at^2 d = 0 + 1/2 (.345) (5.0)^2 = 4.31m

Answer F net = F - mg = ma Since the elevator is moving up the upward force produced by the floor of the elevator is the force produced by gravity added to the force produced by the elevator therefore find the force produced by the elevator on the passenger F = ma + mg = m(a + g) F = 68 kg X (1.0 + 9.8) m/s/s = 730 N From Newton's third law, the passenger exerts a downward force of 730 N on the floor of the elevator.

Answer Since the elevator is accelerating in the same direction as the passenger The apparent weight of the passenger will be less F = m(g - a) F = 68 kg X (9.8 m/s/s - 1.0 m/s/s) = 598.4 N Using Newtons second law: The passenger applies a force of 598.4 N downward on the elevator

Answer Force of elevator on passenger F = mg = 68 kg X 9.8 m/s/s = 670 N Using Newtons third law: Passenger on elevator = 670 N downward

Answer Since it is a constant velocity there is no acceleration Force of elevator on person = mg F= mg = 68 kg X 9.8 m/s/s = 670 N Using Newton's third law Force of passenger on elevator is 670 N downward

Answer Assign up as + The net force in this case is the vector sum of Ftension (positive) and F gravity (negative) Fnet is therefore F tension - F gravity Rearrange Fnet = F tension + F gravity = ma + mg Since vi = 0 a = v^2/2d so: F tension = m(a + g) = m(v^2 / 2d + g) F tension = (50 kg) [(3.0 m/s^2) / 2 (3.0 m) + (9.8 m/s/s)] = 565 N F tension = 565 N, the rope is in danger of breaking because the tension exceeds 500N

Answer Fnormal = F gravity = mg and: Fa = Ff if there is no acceleration therefore: Fa = (mu)(mg) Fa = (0.20)(25 kg)(9.8 m/s/s) = 49 N in the direction of the force

Answer x direction: Fa - Ff = ma y direction: Fn = Fg = mg; Fn = mg Ff  = (mu)(Fn) = (mu)(mg) therefore: a = Fnet / m = Fa - (mu)(mg) /m= (Fa) - (mu)(m)(g) / m so: a = [(98 N ) - (0.20)(25kg)(9.8 m/s/s)] / m = 1.96 m/s/s

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Thanks,

Mr. Clark