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larisa86 [58]
3 years ago
5

A 7kg block slides down a 30 degree slope at a constant velocity of 2 m/s. How big is the force of friction acting on the block?

Physics
1 answer:
lisabon 2012 [21]3 years ago
6 0

The force of friction is <u>34.3 N.</u>

A block of mass m slides down a plane inclined at an angle θ to the horizontal with a constant velocity.  According to Newton's first law of motion, every body continues in its state of rest or a state of uniform motion in a straight line, unless acted upon, by an external unbalanced force. This means that when balanced forces act on a body, the body moves with a constant velocity.

The free body diagram of the sliding block is shown in the attached diagram. Resolve the weight mg of the block into two components  mg sinθ along the direction of the plane and mg cosθ perpendicular to the plane . The force of friction   F acts upwards along the plane  and the normal reaction acts perpendicular to the plane.

Since the block moves down with a constant velocity, the downward force mg sinθ must be equal to the upward frictional force.

F = mg sin\theta

Substitute 7 kg for m, 9.8 m/s² for g and 30° for θ.

F = mg sin\theta = (7 kg)(9.8 m/s^2)(sin30^o) =34.3 N

The force of friction is <u>34.3 N</u> up the plane.


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Answer:

a) c=1822.3214\ J.kg^{-1}.K^{-1}

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

  • mass of aluminium, m_a=0.1\ kg
  • mass of water, m_w=0.25\ kg
  • initial temperature of the system, T_i=10^{\circ}C
  • mass of copper block, m_c=0.1\ kg
  • temperature of copper block, T_c=50^{\circ}C
  • mass of the other block, m=0.07\ kg
  • temperature of the other block, T=100^{\circ}C
  • final equilibrium temperature, T_f=20^{\circ}C

We have,

specific heat of aluminium, c_a=910\ J.kg^{-1}.K^{-1}

specific heat of copper, c_c=390\ J.kg^{-1}.K^{-1}

specific heat of water, c_w=4186\ J.kg^{-1}.K^{-1}

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)

Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)

Q_{in}=11375\ J

The heat released by the blocks when dipped into water:

Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)

where

c= specific heat of the unknown material

For the conservation of energy : Q_{in}=Q_{out}

so,

11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)

c=1822.3214\ J.kg^{-1}.K^{-1}

b)

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c)

The material is peat, possibly.

d)

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

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Answer:

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Answer:

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