/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 A block of metal weighting \(2 \... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 49

A block of metal weighting \(2 \mathrm{~kg}\) is resting on a frictionless plane. It is stuck by a jet releasing water at a rate of \(1 \mathrm{~kg} / \mathrm{sec}\) and at a speed of \(5 \mathrm{~m} / \mathrm{s}\). The initial acceleration of the block will be (A) \(2.5 \mathrm{~m} / \mathrm{s}^{2}\) (B) \(5.0 \mathrm{~m} / \mathrm{s}^{2}\) (C) \(7.5 \mathrm{~m} / \mathrm{s}^{2}\) (D) \(10 \mathrm{~m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
The initial acceleration of the block will be (A) \(2.5 \mathrm{~m} / \mathrm{s}^{2}\).

Step by step solution

01

Calculate the momentum of the water jet

The momentum of the water jet is equal to the mass flow rate multiplied by the speed of the water. The mass flow rate is 1 kg/s, and the speed is 5 m/s. So, the momentum of the water jet can be calculated as follows: Momentum = mass flow rate * speed Momentum = (1 kg/s) * (5 m/s) = \[5 \mathrm{~kg}\cdot\mathrm{m} / \mathrm{s}^2\]
02

Calculate the force exerted by the water jet on the block

To find the force exerted by the water jet on the block, we'll use Newton's Second Law of Motion (F = m*a). Here, the momentum of the water jet works as a force on the block. Since the mass flow rate is constant, the force exerted by the water jet on the block is also constant. Thus, Force = Momentum = \[5 \mathrm{~kg}\cdot\mathrm{m} / \mathrm{s}^2\]
03

Calculate the initial acceleration of the block

Now that we have calculated the force exerted by the water jet on the block, we can use Newton's Second Law of Motion to find the initial acceleration of the 2 kg block: Force = m * a Given Force = \[5 \mathrm{~kg}\cdot\mathrm{m} / \mathrm{s}^2\] Given mass (m) = 2 kg Rearrange the formula to solve for acceleration (a): a = Force / m Substitute the given values: a = (5 kg 鈰� m/s虏) / (2 kg) = \[2.5 \mathrm{~m} / \mathrm{s}^{2}\] Thus, the initial acceleration of the block will be (A) \(2.5 \mathrm{~m} / \mathrm{s}^{2}\).

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Most popular questions from this chapter

A mass \(A\) is released from the top of a frictionless inclined plane \(18 \mathrm{~m}\) long and reaches the bottom 3 second later. At the instant when \(A\) is released, a second mass \(B\) is projected upwards along the plane from the bottom with a certain initial velocity. The mass \(B\) travels a distance up the plane, stops and returns to the bottom simultaneously with \(A\). The two masses do not collide. Initial velocity of \(B\) is (A) \(4 \mathrm{~ms}^{-1}\) (B) \(5 \mathrm{~ms}^{-1}\) (C) \(6 \mathrm{~ms}^{-1}\) (D) \(7 \mathrm{~ms}^{-1}\)

A body of mass \(m_{1}\) makes a head on collision perfectly elastic with a body of mass \(m_{2}\) initially at rest. (i) What fraction of initial energy of mass \(m_{1}\) is lost in collision? (ii) For what ratio of \(\frac{m_{2}}{m_{1}}=\eta\), the fraction of energy loss is maximum?

A particle of mass \(m\) moving in the \(x\)-direction with speed \(2 v\) is hit by another particle of mass \(2 m\) moving in the \(y\) direction with speed \(v\). If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to (A) \(50 \%\) (B) \(56 \%\) (C) \(62 \%\) (D) \(42 \%\)

A particle moves in the \(x-y\) plane under the action of force \(\vec{F}\) such that the value of its linear momentum \(\vec{P}\) at time \(t\) is \(P_{x}=2 \cos t\) and \(P_{y}=2 \sin t .\) The angle \(\theta\) between \(\vec{F}\) and \(\vec{P}\) at time \(t\) will be (A) \(90^{\circ}\) (B) \(0^{\circ}\) (C) \(180^{\circ}\) (D) \(30^{\circ}\)

A ball of mass \(m\) moving at a speed \(u\) makes a head on collision with an identical ball kept at rest. The kinetic energy of the balls after the collision is \(3 / 4^{\text {th }}\) of the original. The co-efficient of restitution will be (A) \(\frac{1}{\sqrt{2}}\) (B) \(\frac{1}{2}\) (C) \(\frac{1}{\sqrt{3}}\) (D) \(\frac{1}{3}\)
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