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hoa [83]
3 years ago
15

A sled is pulled at a constant velocity across a horizontal snow surface. If a force of 100 N is being applied to the sled rope

parallel to the ground, what is the force of friction between the sled and the snow?
Physics
1 answer:
tangare [24]3 years ago
3 0

Answer:

60.18 N

Explanation:

Given that:

The force applied on the sled = 100 N

Suppose, the angle between the sled rope and the ground = 53°

The horizontal force which acts  in the horizontal direction can be expressed as:

F_x = F \ cos \theta

F_x = 100 \ cos (53)

F_x = 60.18 \ N

But if the angle between the sled rope is parallel to the ground. Then, we use an angle on a straight line which is = 180°

F_x = F \ cos \theta

F_x = 100 \ cos (180)

= 100 × -1

= -100 N

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3 years ago
A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.
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Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, KE_{0}  = 0.5 mu^{2}

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, KE_{f} = 0.5mv^{2}

a) Using the principle of energy conservation,

KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

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