Physics, 08.01.2021 01:00 willcohen42

77. Two blocks, with masses m1 = 400 g and m2 = 600 g, are connected by a string and lie on a frictionless tabletop. A force F = 3.5 N is applied to block m2. Find the acceleration of each object.

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Answer from: sam57884

The acceleration of each block connected by the given string is 3.5 m/s².

The given parameters;

mass of the first block, m1 = 400 g = 0.4 kgmass of the second block, m2 = 600 g = 0.6 kgapplied force, F = 3.5 N

The net force on the blocks is calculated as follows;

$\Sigma F = ma\\\\F - T = m_2a\\\\F - m_1(a -g)= m_2a\\\\g = 0 \ , \ in \ horizontal \ direction\\\\F - m_1a = m_2 a\\\\F = m_2 a + m_1 a\\\\F = a(m_1 + m_2)\\\\a = \frac{F}{m_1 + m_2} \\\\a = \frac{3.5 }{0.4 + 0.6} \\\\a = 3.5 \ m/s^2$

Thus, the acceleration of each block connected by the given string is 3.5 m/s².

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77. Two blocks, with masses m1 = 400 g and m2 = 600 g, are connected by a string and lie on a fricti

Answer from: shy88

See the answers below. And the free body diagrams attached.

Explanation:

To solve this problem, we must build a free body diagram, where we show the forces acting on a body, first for the m2 body, where the force of 3.5 [N] acts to the right. In such a way that a reaction force is presented in the opposite direction due to the string tied between both bodies, we will call this force T1.

Now using Newton's second law, which is defined as the sum of forces equal to the product of mass by acceleration, we can determine an equation as a function of T1.

∑F = m*a

$3.5 -T_{1}=0.6*a_{2}$

We have an equation with two unknowns, in such a way we must perform a second free body diagram for the body with mass m1, to acquire the additional equation.

∑F =m₁*a₁

$T_{1}=0.4*a_{1}$

By kinematics we know that if the string tied to Body 2 moves a distance, the part of the string at the end tied to body 1 will move the same distance. This same analysis is valid for velocities and accelerations.

This is the accelerations a1 and a2 are equal. Now equalizing both equations.

$3.5-0.4*a_{1}=0.6*a_{2}\\but\\a_{1}=a_{2}\\3.5-0.4*a_{1}=0.6*a_{1}\\a_{1}=3.5[m/s^{2} ]$

Answer from: Quest

10 is the . . .

Answer from: Quest

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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77. Two blocks, with masses m1 = 400 g and m2 = 600 g, are connected by a string and lie on a fricti...

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