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IRISSAK [1]
3 years ago
14

A rope is used to pull a 4.36 kg block at constant speed 4.33 m along a horizontal floor. The force on the block from the rope i

s 7.97 N and directed 26.2° above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and floor?
Physics
1 answer:
geniusboy [140]3 years ago
4 0

Answer:

Explanation:

The horizontal component of force applied = 7.97 cos 26.2 = 7.15 N.

Vertical component in upward direction = 7.97 sin 26.2 = 3.52 N.

Since the body is moving with uniform velocity, friction force will equalize the external horizontal component = 7.15 N.

So frictional force  = 7.15 N.

a) Work done by rope force per second = force in horizontal direction x displacement per second = 7.15 x 4.33 = 30.95 J

b) Increase in thermal energy per second will be due to negative work done by frictional force = work done by external force = 30.95 J.

c) Normal force acting downwards = weight - vertical component of external force  = 4.36 x 9.8 - 3.52 = 39.21 N

coefficient of friction = friction force / normal force = 7.15 / 39.21 = 0.18

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3 years ago
A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be 118 km
ruslelena [56]

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x component 60.85 m

y component 101.031 m

Explanation:

We have given distance r = 118 km

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3 years ago
2) Two ice skaters have masses m1 and m2 and are initially stationary. Their skates are identical. They push against one another
worty [1.4K]

Answer:

m_1 / m_2 = sqrt (1 / 2)

Explanation:

Given:

- Initial velocity of both skaters V_i = 0

- Velocity of skater 1 after push = V_1

- Velocity of skater  after push = V_2

- Distance traveled by skater 1 = s_1

- Distance traveled by skater 2 = s_2

- s_1 = 2*s_2

- Accelerations of both skaters to halt is equal

Find:

What is the ratio m1/m2 of their masses

Solution:

- Apply conservation of momentum for two skaters just before and after the push as follows:

                                              P_i = P_f

                                  0 = m_1*V_1 - m_2*V_2

- Evaluate:                 m_1 / m_2 = ( V_2 / V_1 )

- Apply Conservation of Energy on both skaters as follows:

- Skater 1:

                               0.5*m_1*V_1^2 = u_k*m_1*g*s_1

-Simplify:                      0.5*V_1^2 = u_k*g*(2*s_2)

- Skater 2:

                               0.5*m_2*V_2^2 = u_k*m_2*g*s_2

-Simplify:                      0.5*V_2^2 = u_k*g*s_2

- Divide the two energy equations for skaters:

                                    (V_1 / V_2)^2 = 2

                                    (V_2 / V_1)^2 = 1 / 2

- simplify:                     (V_2 / V_1) = sqrt (1 / 2)

-Hence from earlier momentum conservation results:

                                  m_1 / m_2 = ( V_2 / V_1 ) = sqrt (1 / 2)

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