Question

A 2.0-kg block is held at rest against a spring of constant 2700 N/m, compressing it 5.5 cm. The block and the spring are located on a horizontal surface. When the system is released, the block slides down a ramp with an angle of 35° to a second horizontal surface 1.7 m below the first. The incline has a coefficient of kinetic friction of 0.29; the horizontal surfaces are frictionless. Please find a) the potential energy stored in the spring before it is released. b) the mechanical energy lost by the block once it reaches the lower horizontal surface. c) the final speed of the block as it slides along the lower horizontal surface.

Answer 1

Answer:

a) 4.1 J

b) -14 J

c) 4.8 m/s

Explanation:

The energy stored in the spring is given by:

$U_e=\frac{1}{2}*K*x^2\\\\U_e=\frac{1}{2}*2700N/m*(5.5*10^{-2}m)^2\\\\U_e=4.1J$

The mechanical energy loss is because of the work done by the friction force.

The friction force (only presented on the inclined surface) is given by:

$F_f=\µ*F_N$

$F_N=2.0kg*9.8m/s^2*cos(35)\\F_N=16N\\F_f=0.29*16N=4.6N$

We need to calculate the length of the ramp in order to calculate the work, the length of the ramp is the hypotenuse:

$sin(\theta)=\frac{OC}{h}\\h=\frac{OC}{sin(\theta)}\\\\h=\frac{1.7m}{sin(35)}\\\\h=3.0m$

So the work done by the friction force is:

$W_f=F_f*d*cos(\alpha)\\W_f=4.6N*3.0*cos(180)\\W_f=-14J$

the angle is 180 degrees because the force is opposite to the motion.

In order the know the final velocity we need to apply the Energy Conservation Theorem:

$K_1+U_{g1}+U_{e}+W_f=K_2+U_{g2}\\\\0+m*g*h_1+4.1J-14J=\frac{1}{2}*m*v^2+m*g*h_2\\\\2.0kg*9.8m/s^2*1.7m+4.1J-14J=\frac{1}{2}*2.0kg*v^2+2.0kg*9.8m/s^2*(0)\\\\33J+4.1J-14J=v^2\\\\v=\sqrt{23.1}\\\\v=4.8m/s$