# Mechanics Homework, Set 6

1. Unless otherwise stated, all objects are located near the Earth’s surface, where $g= 9.80 m/s^2$.

A rocket is far away from all planets and stars, so gravity is not a consideration. It is using its rocket engines to accelerate upward with an acceleration $a = 9.80 m/s^2$. On the floor of the main deck is a crate (object with brick pattern) with a mass of 71.0 kg. A) How many forces are acting on the crate? B) Determine the normal force on the crate. C) Compare normal force on the crate to the normal force the crate would experience if it were at rest on the surface of the Earth.

2. all objects are located near the Earth’s surface, where $g = 9.80 m/s^2$. Assuming ideal frictionless conditions for the apparatus shown in the figure , what is the acceleration of the system in the following conditions. A) m1= 0.26 kg, m2 = 0.51 kg, and m3= 0.26 kg. Find the magnitude of the acceleration. Express your answer using two significant figures. B) Find the direction of the acceleration.Express your answer using two significant figures. C) m1= 0.21 kg, m2=0.16 kg, and m3= 0.51 kg. Find the magnitude of the acceleration? D)Find the direction of the acceleration.

3. The floor of the basement of a house is 2.8 m below ground level, and the floor of the attic is 4.6 m above ground level. A)If an object in the attic were brought to the basement, the change in potential energy will be greatest relative to which floor? B) What are the respective potential energies of 2.3 kg objects in the basement and attic, relative to ground level? Express your answers using two significant figures separated by a comma. C) What is the change in potential energy if the object in the attic is brought to the basement?

4.A 2.5 kg box that is sliding on a frictionless surface with a speed of 13 m/s approaches a horizontal spring. (see the figure ) The spring has a spring constant of 2100 n/m. A) If one end of the spring is fixed and the other end changes its position, how far will the spring be compressed in stopping the box? B)How far will the spring be compressed when the box’s speed is reduced to half of its initial speed?

5. A sleigh and driver with a total mass of 130 kg are pulled up a hill with a 15 degrees incline by a horse, as illustrated in the figure. A)If the overall retarding frictional force is 850 N and the sled moves up the hill with a constant velocity of 5.0 km/h , what is the power output of the horse? (Express in horsepower, of course.) B)Suppose that in a spurt of energy, the horse accelerates the sled uniformly from 5.0 km/h to 18 km/h in 7.0 s . What is the horse’s maximum instantaneous power output? Assume the same force of friction. Express your answer using two significant figures.

On the brick act two forces : the first one is the apparent weight (or equivalent its inertia) $Ga= m*a$ directed downwards and the reaction force of the floor $N=-Ga$, directed upwards.

B)

The Normal force depends on what is defined as ‘normal’. One can understand that the normal force is the reaction force from the floor $N =-Ga =-m*a =-71*9.80 =-695.8 N$ or normal force (in the perpendicular direction to the brick) can be null because the action (Ga) is equal to the reaction (from the floor) as principle third states.

C)

The normal force on the spaceship is equal to the normal force at rest on the surface of the earth because the gravitational acceleration g is equal to the acceleration of the rocket.

$m2 > m1$ , the system formed by the three masses moves to the right, m2 moves down. (the positive axe is to the right, m2 down, m1 up)

$(m1+m2+m3)*a = m2*g -m1*g$

$a = g(m2-m1)/(m1+m2+m3) =9.8*(0.51-0.26)/(0.26+0.51+0.26) =2.379 m/s^2$

B)

acceleration is in the direction of the positive axe

C)

$m1 > m2$, the system moves to the left, m1 down, m2 up. the positive axe is to the left , m1 down, m2 up.

$(m1+m2+m3)*a =(m1-m2)*g$

$a =g(m1-m2)/(m1+m2+m3) =9.8*(0.21-0.16)/(0.21+0.16+0.51) =0.557 m/s^2$

D)

acceleration is in the direction of positive axe.

If the object is brought from the attic to the basement, the change in potential energy is greatest (in absolute value) relative to the attic floor. (the attic floor is taken as zero level for the potential energy). In this case the potential energy of the object on the basement floor is

$Ep = -m*g*(2.8+4.6)$

B)

Relative to the ground level (the ground level is taken as zero potential energy) the potential energy of an object on the attic is $E1 = m*g*4.6 =2.3*9.8*4.6 = 103.684 J$ and the potential energy of an object in the basement is $E2 =-m*g*2.8 =-2.3*9.8*2.8 =-63.112 J$

C)

If the object in the attic is brought to the basement the change in the potential energy is

$\Delta(E) = E(basement) -E(attic) = 0 -m*g*(2.8+4.6) =-2.3*9.8*(2.8+4.6) =$

$=-166.796 J$

4. A)

The total energy of the system box + spring is the same initial and final. Initial the energy is kinetic (box energy) $Ec = (m*v^2)/2$ Final energy is potential (compressed string energy) $Ep =(k*x^2)/2$ where x is the compression distance of the spring.

$(mv^2)/2 =(k*x^2)/2$

$x = v*\sqrt{m/k} =13*\sqrt{2.5/2100} =0.448 m$

B)

again the total energy is the same . initial speed this time v = 13/2 =6.5 m/s

$(mv^2)/2 =(k*x^2)/2$

$x = v*\sqrt{m/k} =6.5*\sqrt{2.5/2100} =0.224 m$

5. A)

$Power = Force*speed$

The force of the horse is equal to the friction force since the speed is the same.

$5 Km/h = 5000/3600 =1.389 m/s$

$P= F*V= 850*1.389 =1180.56 W =1180.56/735.5 =1.6 HP$

B)

$18 km/h = 18000/3600 =5 m/s$

the acceleration is $a= (V2-V1)/t =(5-1.389)/7 =0.516 m/s^2$

The force on the sleigh is the friction force plus its inertia

$F = Ff+m*a =850+130*0.516 =917.08 N$

The maximum power output of the horse is

$P_{max} =F*V_{max} =917.08*5 =4585.40 W = 6.23 HP$