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Alex_Xolod [135]
3 years ago
7

A 1500 kg car begins sliding down a 5.0o inclined road with a speed of 30 km/h. The engine is turned off, and the only forces ac

ting on the car are a net frictional force from the road and the gravitational force. After the car has traveled 50 m along the road, its speed is 40 km/h.
(a) How much is the mechanical energy of the car reduced because of the net frictional force?
(b) What is the magnitude of that net frictional force?
Physics
1 answer:
Marina86 [1]3 years ago
4 0

Answer:

a) -23583J

b) 471.66N

Explanation:

Using the conservation energy theorem:

K_o+U_o+W_f=K_f+U_f\\W_f=K_f+U_f-(K_o+U_o)

The height is given by:

sin(\theta)=\frac{h}{d}\\\\h=50*sin(5^o)\\h=4.36m

substituting the values we have:

W_f=\frac{1}{2}*1500kg*((40-30)\frac{km}{h}*\frac{1000h.m}{3600km.s})^2+0-1500kg*9.8m/s^2*4.36m\\W_f=-23583J*

the work is defined by:

W_f=F_f*d*cos(\theta)

the force of the friction force is 180 degrees because it's opposite to the movement, so:

F_f=\frac{-23583J}{50m*cos(180)}\\F_f=471.66N

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Answer:

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

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Where:

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If we know that \vec r_{cm} = (-0.500\,m,-0.700\,m), m_{1} = 1\,kg, \vec r_{1} = (-1.20\,m, 0.500\,m), m_{2} = 4.50\,kg, \vec r_{2} = (0.600\,m, -0.750\,m) and m_{3} = 4\,kg, then the coordinates of the third particle are:

(-0.500\,m, -0.700\,m) = \frac{(1\,kg)\cdot (-1.20\,m,0.500\,m)+(4.50\,kg)\cdot (0.600\,m,-0.750\,m)+(4\,kg)\cdot \vec r_{3}}{1\,kg+4.50\,kg+4\,kg}

(-4.75\,kg\cdot m, -6.65\,kg\cdot m) = (-1.20\,kg\cdot m, 0.500\,kg\cdot m) + (2.7\,kg\cdot m, -3.375\,kg\cdot m) +(4\cdot x_{3},4\cdot y_{3})

(4\cdot x_{3}, 4\cdot y_{3}) = (-6.25\,kg\cdot m,-3.775\,kg\cdot m)

(x_{3},y_{3}) = (-1.562\,m,-0.944\,m)

a) The x coordinate of the third mass is -1.562 meters.

b) The y coordinate of the third mass is -0.944 meters.

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