6. The vertical distance of an elevator above the ground floor is given by y = 0

Question

6. The vertical distance of an elevator above the ground floor is given by y = 0 4 t^2 (where lengths are in meters and time is in seconds). What is the normal force of the elevator floor on an 85 kg person riding inside? (HINT: First find the acceleration of the elevator.) a) 795 N b) 848 N c) 872 N d) 901 N e) 954 N 7. A bob at the end of a thin wire of length L = 50 cm is rotated in a horizontal circle with a constant speed v. The suspending wire traces out a cone with an angle of 30 degree with the vertical direction. Find the speed of the bob. a) 0.982 m/s b) 1.19 m/s c) 1.40 m/s d) 1.63 m/s e) 0.833 m/s 8. A block moves up a 20 degree incline with constant speed under the action of a force of 15 N applied parallel to the incline. If the coefficient of kinetic friction is 0.3, what is the crass of the block? a) 3.77 kg b) 3.43 kg c) 3.11 kg d) 2.78 kg e) 2.45 kg 9. A waitress shoves a ketchup bottle to her right along a horizontal lunch counter. The bottle leaves her hand moving at 2.0 m/s and then slows down as it slides because of a constant horizontal friction force. The coefficient of kinetic friction between the bottle and the counter is 0.53. How far does it slide before coming to rest? a) 0.466 m b) 0.425 m c) 0.509 m d) 0.263 m e) 0.385 m 10. A wagon with mass 7 kg moves in a straight line on a frictionless horizontal surface. It has an initial speed of 4.00 m/s and then is pushed 2.0 m in the same direction as the velocity by a force with magnitude of 10 N Find the wagon's final speed a) 5.10 m/s b) 4.66 m/s c) 4.96 m/s d) 4.81 m/s e) 5.63 m/s

Explanation / Answer

6) y = 1/2 a t^2

so 1/2 a = 0.4

a = 0.8

N - m g = m a

N = ma + mg = 85*(9.81+0.8)= 901 N

6)

sum forces in the y

T cos 30- m g = 0

T = mg/cos 30

sum in the x

T sin 30 = mv^2/r

m g tan 30 = m v^2/(L sin 30)

9.81*tan(30) = v^2/(.5*sin(30))

v=1.19 m/s

8) so sum forces parallel to incline

F - m g sin theta - friction = 0

friction = u N = u m g cos theta

15 - m*9.81*sin(20 degrees) - 0.3*m*9.81*cos(20 degrees)=0

m=2.45 kg

9) a = u g

v^2 = v0^2 - 2 a x

0^2 = 2^2 -2*0.53*9.81*x

solve for x

10)

W = dKE

W = F d

10*2 = 0.5*7*(v^2 - 4^2)

solve for v

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