A 10-kg crate resting on a horizontal surface is pulled with a force of 60N at a

Question

A 10-kg crate resting on a horizontal surface is pulled with a force of 60N at a 40.0° angle to the horizontal What is the normal force acting on the crate? 1. T=60 N 40° 10 kg a. 137 N b. 98 N c. 59 N d. 52 N 2. A 1000 kg car is parked along a 15° incline. The normal force acting on the car is a. 9800N b. 9466 N _ c. 2536 N d. 259 N 3. A block remains stationary on an incline of 30°, but slides down when the angle is increased. The coefficient of friction between the two surfaces is a. Kinetic, 0.50 b. Kinetic, 0.58 c. Static, 0.50 d. Static, 0.58 A person is standing on a bathroom scale in an elevator that is slowing down as it moves upwards. The weight displayed on the scale will beher actual weight. a. Less than b. More than 4. c. The same as 5. A block is resting on an incline. The correct directions of the normal force and gravity are shown in Fig_ c. C d. D b. B sphere of radius r falling vertically down with speed v experiences a drag force of D. If its radius doubles and speed becomes half, the drag will be: a. D b. 2D 6. A c. D/2 d. D/4

Explanation / Answer

1. F_net y = Fn + T sin40 - m g = 0

Fn = (10 x 9.8) - 60 sin40 = 59.4 N  

Ans(c)

2. N = m g cos(theta)

N = 1000 x 9.8 x cos15 = 9466 N  

Ans(b)

3. it just starts to slide, before it was stationary.

static fricion.

u = tan(theta) = tan30 = 0.58  

Ans(d)

4. a -> downward

Fn = m ( g- a)

Ans(a) less than

5. Ans(B)

6. Fd = rho A Cd v^2 / 2 = D

r' = 2 r then A' = 4 A

v' = v/2 then v'^2 = v / 4

then Fd = rho (4 A)(Cd)(v^2/ 4) /2 = rho A Cd v^2 /2 = D

Ans(A)

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