Two 5-N boxes are attached to opposite ends of a spring scale and numbered from
ID: 1414520 • Letter: T
Question
Two 5-N boxes are attached to opposite ends of a spring scale and numbered from pulleys as shown. What is the reading on the scale? (a) 0N (b) 2.5 N (c) 5N (d) 10N (e)25 N Suppose that the system were placed in an elevator that accelerated downward as 2 m/s^2. (a) 0N (b) 4N (c) 6N (d) 8N (c) 10N A rope holds a 10-kg rock at rest on a frictionless inclined plane as shown. Determine the tension in the rope. (a) 9.8 N (b) 20N (c) 49N (d) 85N (e) 98N Which one of the following statements concerning the force exerted on the plane by the rock is true? (a) It is 0 N. (b) It is 98 N. (c) It is greater than 98 N. (d) It is less than 98 N, but greater than zero newtons. (e) It increases as angle inclination is increased. Determine the magnitude of the acceleration of the rock down the inclined plane if the rope breaks? (a) zero m/s^2 (b) 4.9 m/s^2 (c) 5.7 m/s^2 (d) 8.5 m/s^2 (e) 9.8 m/s^2 At playground, a child slides down a slide that makes a 42degree angle with the horizontal direction. The coefficient of kinetic friction for the child sliding on the slide is 0.20. What is the magnitude of her acceleration during her sliding? (a) 4.6 m/s^2 (b) 5.1 m/s^2 (c0 5.4 m/s^2 (d) 6.3 m/s^2 (e) 9.8 m/s^2Explanation / Answer
ANSWER
21.
Each string exerts a force of 5 N to suspend the hanging mass to which it is attached. This means the tension in the string is 5 N. So the scale will read 5 N (Option c).
22.
Because the Spring Scale reads only half of each box, on the elevator the scale will read 4N (Option b).
23.
Fg = 10*9.8 = 98 N
Fgx = Fg*Sin30 = 98*Sin30 = 49 N (Option c).
24.
Because of Fn = 84.9 N. The correct answer is It is less than 98 N, but greater than zero newtons (Option d).
25.
Acceleration = g*sin 30 = 9.8*1/2 = 4.9m/s2 (Option b).
26.
The angle of the slide theta = 42º
the mass of the child is m
when she slides down now apply Newton's law
ma = mgsin(theta) - f
ma = mgsin(theta) - µmgcos(theta)
therefore the acceleration
a = gsin(theta) - µ gcos(theta)
= g [ sin(theta) - µ cos(theta)]
= (9.8m/s^2 )[ sin42 - (0.2)cos42]
= 5.1 m/s^2 (Option b).
Regards!!!
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