Question

an object with initial velocity v0, as shown above, slides up and then down a long, frictionless, inclined plane. which of the following is true of the object as it moves? (a) it has a constant acceleration while moving up the plane and a greater acceleration when moving down the plane. (b) it has a constant acceleration while moving up the plane and a smaller acceleration when moving down the plane. (c) it moves with a constant velocity both up and down the plane. (d) it has the same acceleration as it moves up and down the plane. (e) it has a continually varying acceleration as it moves up and down the plane. 3. an object weighing 300 n is suspended by means of two cords, as shown above. the tension in the horizontal cord is (a) 0 n (b) 150 n (c) 210 n (d) 300 n (e) 400 n 4. a force f accelerates a system of two blocks, x and y, on a horizontal frictionless surface, as shown above. the acceleration is 4 m/s². the force with which block y pushes on block x has magnitude

Explanation:

Step1: Analyze forces on inclined - plane motion

For an object on a frictionless inclined plane, the net - force along the plane is $F = mg\sin\theta$, where $m$ is the mass of the object, $g$ is the acceleration due to gravity, and $\theta$ is the angle of the inclined plane. According to Newton's second law $F = ma$, so $a = g\sin\theta$. The acceleration is constant both up and down the plane because $g$ and $\theta$ are constant, and the direction of acceleration is down - the - plane both when the object is moving up (it is decelerating) and when it is moving down (it is accelerating).

Step2: Analyze forces on suspended object

Let the tension in the horizontal cord be $T_1$ and the tension in the angled cord be $T_2$. Resolve the forces in the vertical and horizontal directions. In the vertical direction, $T_2\sin45^{\circ}=mg = 300N$ (since the object is in equilibrium vertically). In the horizontal direction, $T_1=T_2\cos45^{\circ}$. From $T_2\sin45^{\circ}=300N$, we can find $T_2=\frac{300}{\sin45^{\circ}}=\frac{300}{\frac{\sqrt{2}}{2}} = 300\sqrt{2}N$. Then $T_1=T_2\cos45^{\circ}=300N$.

Step3: Analyze forces on two - block system

The total mass of the system $m_{total}=m_X + m_Y=2kg + 4kg = 6kg$. The acceleration of the system $a = 4m/s^2$. According to Newton's second law, the net force on the system $F = m_{total}a=6\times4 = 24N$. The force exerted by block $Y$ on block $X$ is the force required to accelerate block $Y$ only. Using Newton's second law $F_{Y - on - X}=m_Ya$. Substituting $m_Y = 4kg$ and $a = 4m/s^2$, we get $F_{Y - on - X}=16N$. But we are not asked for this value in the given sub - questions.

Answer:

1. For the inclined - plane motion question: D. It has the same acceleration as it moves up and down the plane.
2. For the suspended object question: D. 300 N