$\bar a=\dfrac{\Delta v}{\Delta t}=\dfrac{0\,\frac{\mathrm m}{\mathrm s}-20\,\frac{\mathrm m}{\mathrm s}}{3.5\,\mathrm s-0\,\mathrm s}=-5.7\,\dfrac{\mathrm m}{\mathrm s^2}$

so the answer is D.

6 0

2 years ago

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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 locate

KonstantinChe [14]

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:

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

5 0

2 years ago