What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0190 m). See details.

A 2.55 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0390 m. The spring has a force constant of 815 N/m. The coefficient of kinetic friction between the floor and the block is 0.35. The block and spring are released from rest and the block slides along the floor.

Answer 1

#v=0.32" "m/s#

#"When a spring is compressed by a Force of F,The potential energy "##"is stored on it."#

#"if the spring is released, the Potential energy turns into kinetic energy."#

#"The object is repulsed by the call-back force which occurs on the spring"#

#F'=K*Delta x#

#K=815 N/m#
#Delta x=0.0190" "m#

#F'=815*0.019=15.48" "N#

#F_f=k*m*g#

#F_f=0.35*2.55*9.81=8.76N" Friction force"#

#"acceleration of object is :"#

#a=(F'-F_f)/m#

#m=2.55kg" mas of object"#

#a=(15.48-8.76)/(2.55)#

#a=2.64" "m/s^2#

#v^2=2*a*displacement#

#v^2=2*2.64*0.0200#

#v^2=0.1056#

#v=sqrt(0.1056)#

#v=0.32" "m/s#

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Answer 2

To find the speed of the block when it has moved a distance of 0.0200 m from its initial position, we can use the principle of conservation of mechanical energy. The initial potential energy stored in the compressed spring is equal to the final kinetic energy of the block.

Let's denote the initial compression of the spring as x, the spring constant as k, and the mass of the block as m. Using Hooke's Law, we have:

(F = -kx)

Where F is the force exerted by the spring.

The potential energy stored in the spring when it is compressed by a distance x is given by:

(PE = \frac{1}{2} kx^2)

When the block has moved a distance of 0.0200 m from its initial position, the spring is compressed by 0.0190 m (since the spring is compressed by x when the block moves by 0.0200 m). So, we can calculate the initial potential energy stored in the spring using the value of x.

Next, we equate this potential energy to the kinetic energy of the block:

(KE = \frac{1}{2} mv^2)

Where v is the speed of the block.

By setting the initial potential energy equal to the final kinetic energy, we can solve for v.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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