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3 years ago

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 34.3 N.

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 34.3 N up the plane.

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An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10

Alexeev081 [22]

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,

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

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

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).

The material is peat, possibly.

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

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2 years ago

A projectile is launched at an angle of 60 degrees to the horizontal at an initial speed of 10 meters per second. What is the ma

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Answer:

5 sq. root 3

Explanation:

theta= 60°

=> u sin theta = 10 × sin 60

= 10× sq. root 3/2

= 5 sq. root 3

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3 years ago

The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give

Flauer [41]

Answer:

velocity = 1527.52 ft/s

Acceleration = 80.13 ft/s²

Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700(0.019)²)/sin66

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Answer:

A theory is an explanation for what has been shown many times. A scientific law is a relationship in nature that has been proved many times and there are no exceptions.

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