Problem 7.53 A 0.400 kg potato is tied to a string with length 1.90 m , and the
ID: 1548001 • Letter: P
Question
Problem 7.53
A 0.400 kg potato is tied to a string with length 1.90 m , and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released
Part B
What is the tension in the string at this point? T= N
Exercise 7.33
A small block with mass 0.0400 kg is moving in the xy-plane. The net force on the block is described by the potential- energy function U(x,y)= (5.50 J/m2 )x2-(3.85 J/m3 )y3.
Part A
What is the magnitude of the acceleration of the block when it is at the point x= 0.35 m , y= 0.52 m ?
Express your answer with the appropriate units.
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Part B
What is the direction of the acceleration of the block when it is at the point x= 0.35 m , y= 0.52 m ?
Exercise 7.33
A small block with mass 0.0400 kg is moving in the xy-plane. The net force on the block is described by the potential- energy function U(x,y)= (5.50 J/m2 )x2-(3.85 J/m3 )y3.
Part A
What is the magnitude of the acceleration of the block when it is at the point x= 0.35 m , y= 0.52 m ?
Express your answer with the appropriate units.
a=SubmitMy AnswersGive Up
Part B
What is the direction of the acceleration of the block when it is at the point x= 0.35 m , y= 0.52 m ?
= counterclockwise from the +x-axisExplanation / Answer
Here
m = 0.400 Kg
L = 1.90 m
part B) at the bottom point
velocity of the object , v = sqrt(2 * 9.80 * 1.90)
NOw, for the tension in the string
tension in the string = m * v^2/L + m * g
tension in the string = 0.40 * (2 * 9.80 * 1.90/1.90 + 9.80)
tension in the string = 11.76 N
the tension in the string at this point is 11.76 N