Question

Item 1

The figure presents a block on a horizontal surface attached to a horizontal spring. The other end of the spring is attached to a wall. The spring is labeled 50 newtons per meter, and the box is labeled 0.5 kilogram. The horizontal position of the center of spring is x equals negative 0.3 meter. The horizontal position of the center of the box is x equals 0 meters. When the spring is stretched the box will be at a horizontal position of x equals 0.3 meter.
A block of mass 0.5kg on a horizontal surface is attached to a horizontal spring of negligible mass and spring constant 50N/m. The other end of the spring is attached to a wall, and there is negligible friction between the block and the horizontal surface. When the spring is unstretched, the block is located at x=0m. The block is then pulled to x=0.3m and released from rest so that the block-spring system oscillates between x=−0.3m and x=0.3m. What is the magnitude of the acceleration of the block and the direction of the net force exerted on the block when it is located at x=0.3m?


Magnitude of Acceleration Direction of Net Force
30m/s2 Positive
, Magnitude of Acceleration Direction of Net Force 30 meters per second squared Positive

Magnitude of Acceleration Direction of Net Force
30m/s2 Negative
, Magnitude of Acceleration Direction of Net Force 30 meters per second squared Negative

Magnitude of Acceleration Direction of Net Force
0 m/s2 Positive
, Magnitude of Acceleration Direction of Net Force 0 meters per second squared Positive

Magnitude of Acceleration Direction of Net Force
0 m/s2 Negative

Answer 1

Step-by-step explanation:

No need to calculate the force you can just find by Hookes law

Hookes law states that

• F=-kx

Where

• F is force ,k is spring constant x is distance

So

force remains negative as per it

• Acceleration=50/0.5÷1/3=100/3=33≈-30m/s²

Option B