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ANEK [815]
1 year ago
12

You push downward on a box at an angle 25° below the horizontal with a force of 750 N. If the box is on a flat horizontal surfac

e for which the coefficient of static friction with the box is 0.8, what is the mass of the heaviest box you will be able to move?
Physics
1 answer:
RoseWind [281]1 year ago
8 0

Answer:

103.44g

Explanation:

R=f/u

=750/0.8

=937.5N

m = 937.5/10cos25

m = 103.44g

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you dribble a basketball while walking on a basketball court list the pairs of action reaction forces in this situation
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Can someone help me with the physics problem?:
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3 years ago
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
3 years ago
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