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3 years ago

A block has two strings attached to it on opposite ends. One string has a force of 5 N,

1 answer:

4 0

Unless you have a diagram to include or any other additional info, I'll assume the block is being pulled by two opposing forces along the horizontal surface.

Horizontally, the block is under the influence of

• one rope pulling in one direction with magnitude 15 N,

• the other rope pulling in the opposite direction with mag. 5 N, and

• friction, opposing the direction of the block's motion, with mag. 3 N.

It stands to reason that the block is accelerating in the direction of the larger pulling force.

(A) By Newton's second law, we have

15 N + (-5 N) + (-3 N) = m (1 m/s²)

where m is the mass of the block. Solve for m :

7 N = m (1 m/s²)

m = (7 N) / (1 m/s²)

m = 7 kg

(B) The friction force is proportional to the normal force, so that if f is the mag. of friction and n is the mag. of the normal force, then f = µ n where µ is the coefficient of friction.

The block does not bounce up and down, so its vertical forces are balanced, which means the normal force and the block's weight (mag. w) cancel out:

n + (-w) = 0

n = w

n = m g

where g = 9.8 m/s² is the mag. of the acceleration due to gravity.

n = (7 kg) (9.8 m/s²)

n = 68.6 N

Then

3 N = µ (68.6 N)

µ = (3 N) / (68.6 N)

µ ≈ 0.044

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3 years ago

An elaborate pulley consists of four identical balls at the ends of spokes extending out from a rotating drum. A box is connecte

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Answer:

its speed will be less than V

Explanation:

When the ball falls a distance d, its final kinetic energy plus rotational kinetic energy of the drum equals its initial potential energy.

K = U

With its speed V at the end of d, we have

1/2mV² + 1/2Iω² = mgd where I = rotational inertia of drum and balls, ω = angular speed of drum and balls and m = mass of box

1/2mV² + 1/2Iω² = mgd

1/2mV² = mgd - 1/2Iω²

V² = [2(mgd - 1/2Iω²)/m]

V = √[2(mgd - 1/2Iω²)/m]

When the four balls are moved inward closer to the drum, their rotational inertia increases and also its angular speed which thus causes an increase in rotational kinetic energy. But, since the box still falls the same distance of d, its final kinetic energy plus rotational kinetic energy of the drum plus balls still equals its initial potential energy

K = U

I' = new rotational inertia of drum and balls, ω' = new angular speed of drum and balls

With its new speed is now V' at the end of d,

1/2mV'² + 1/2I'ω'² = mgd

1/2mV'² = mgd - 1/2I'ω'²

V² = [2(mgd - 1/2I'ω'²)/m]

V' = √[2(mgd - 1/2I'ω'²)/m]

Since I' and ω' increase, the rotational kinetic energy of the drum and balls (1/2I'ω'²) increases. Thus, the difference (mgd - 1/2I'ω'²) < (mgd - 1/2Iω²) which implies that the kinetic energy of the box decreases. Hence, since its kinetic energy decreases, its speed V' also decreases.

So, V' < V

So, its speed will be less than V.

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3 years ago

How much ice (at 0Â°C) must be added to 1.90 kg of water at 79 Â°C so as to end up with all liquid at 8 Â°C? (ci = 2000 J/(kg.Â°

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