Asked by Anonymous
A 5.00-kg block is placed on top of a 10.0-kg block which rests on a horizontal surface. The 5.00-kg block is tied to a
wall with a horizontal string. A 45.0-N horizontal force directed away from the wall is then exerted on the 10.0-kg block.
The coefficient of kinetic friction between the 10.0-kg block and the horizontal surface is 0.100. The coefficient of
kinetic friction between the two blocks is 0.200. (a) Draw the free-body diagram for each block. (b) Determine the
acceleration of the 10.0-kg block. (c) Determine the tension in the string.
I keep getting the wrong answers for this and it's really frustrating me
wall with a horizontal string. A 45.0-N horizontal force directed away from the wall is then exerted on the 10.0-kg block.
The coefficient of kinetic friction between the 10.0-kg block and the horizontal surface is 0.100. The coefficient of
kinetic friction between the two blocks is 0.200. (a) Draw the free-body diagram for each block. (b) Determine the
acceleration of the 10.0-kg block. (c) Determine the tension in the string.
I keep getting the wrong answers for this and it's really frustrating me
Answers
Answered by
Damon
normal forces
5 kg block:
5(9.81)
10 kg block
5*9.81
and
15*9.81
forces
5 kg block
.2 * 5 * 9.81 right
T left
no acceleration so
T = .2 * 5 * 9.81 = 9.81 N
(answer to part C by the way)
10 kg block
T left
.1 * 15 * 9.81 left
45 N right
so
45 - 9.81 - .1*15*9.81 = 10 a
the end
5 kg block:
5(9.81)
10 kg block
5*9.81
and
15*9.81
forces
5 kg block
.2 * 5 * 9.81 right
T left
no acceleration so
T = .2 * 5 * 9.81 = 9.81 N
(answer to part C by the way)
10 kg block
T left
.1 * 15 * 9.81 left
45 N right
so
45 - 9.81 - .1*15*9.81 = 10 a
the end
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