Question 1: A small block with mass 0.0400 kg is moving in the xy -plane. The ne

Question

Question 1:

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)= (6.00 J/m2 )x2-(3.90 J/m3 )y3.

Part A

What is the magnitude of the acceleration of the block when it is at the point x= 0.36 m , y= 0.53 m ?

Express your answer with the appropriate units.

136 ms2 ****ANSWERED****

Part B

What is the direction of the acceleration of the block when it is at the point x= 0.36 m , y= 0.53 m ?

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Question 2:

Two blocks with different mass are attached to either end of a light rope that passes over a light, frictionless pulley that is suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended a distance 1.10 m , its speed is 4.00 m/s.

Part A

If the total mass of the two blocks is 15.0 kg , what is the mass of the more massive block?

Take free fall acceleration to be 9.80 m/s2 .

Part B

What is the mass of the lighter block?

Take free fall acceleration to be 9.80 m/s2 .

136 ms2 ****ANSWERED****

Explanation / Answer

part B:

as we know,

force in x direction=Fx=-dU/dx=-12*x

force in y direction=Fy=-dU/dy=11.7*y^2

(here derivatives are partial derivatives)

at x=0.36 m and y=0.53 m:

Fx=-12*0.36=-4.32 N

Fy=11.7*0.53^2=3.2865 N

hence the force lies in 2nd quadrant and angle with +ve x axis is given by

arctan(Fy/Fx)=arctan(3.2865/-4.32)=142.737 degree (in anticlockwise direction)

as force=mass*acceleration

acceleration will have same direction as force

hence direction of acceleration is 142.737 degree with +ve x axis in anticlockwise direction.

question 2:

part A:

let mass of larger block is m and smaller block is 15-m kg.

then as larger mass descends 1.1 m, the smaller mass will climb 1.1 m from the initial point.

then decrease in potential energy=m*9.8*1.1-(15-m)*9.8*1.1=21.56*m-161.7 J

both the blocks will have smae speed of 4 m/s

then increase in kinetic energy=0.5*m*4^2+0.5*(15-m)*4^2

=120 J

as there is no external force, total energy will be conserved.

hence rise in kinetic energy=decrease in potential energy

==>120=21.56*m-161.7

==>m=(120+161.7)/21.56=13.066 kg

hence mass of larger block is 13.066 kg.

part B:

mass of lighter block=15-m=15-13.066=1.934 kg

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