A rope is tied to a box and used to pull the box 1.5 m along a h zontal floor. The rope makes a

A rope is tied to a box and used to pull the box 1.5 m...

Key Concepts

Gravitational Force

Gravitational force is the force with which the Earth attracts an object towards its center and acts vertically downward. When an object moves horizontally, the gravitational force does no work since there is no displacement in the direction of the force, highlighting the importance of the direction of force relative to displacement in work calculations.

Tension Force

Tension is the pulling force transmitted along a string, rope, or cable. When a force is applied at an angle, tension must be decomposed into its horizontal and vertical components to determine how much of the force contributes to the displacement. The horizontal component is particularly important for calculating work when the motion of the object is along a horizontal surface.

Normal Force

The normal force is the support force exerted by a surface, acting perpendicular to that surface. In many situations, such as when an object moves horizontally, the normal force does no work because it does not have any component in the direction of the displacement.

Friction

Friction is a resistive force that opposes the motion of objects sliding against each other. It usually acts opposite to the direction of displacement and therefore does negative work, removing kinetic energy from the system and often converting it into thermal energy.

Force Decomposition

Force decomposition refers to breaking a force into its perpendicular components, typically using trigonometric functions. This is essential when a force is applied at an angle, as it allows one to distinguish between the components that cause movement along different directions, such as horizontal and vertical, which is particularly useful when analyzing the work done by each component.

Dot Product

The dot product is a mathematical operation that multiplies two vectors and sums the products of their respective components. In physics, it is used to calculate work, as it combines the magnitudes of the force and displacement with the cosine of the angle between them, thereby isolating the effective component of the force that contributes to movement.

Work

Work is the energy transferred when a force causes an object to be displaced. It is calculated as the dot product of the force vector and the displacement vector, which incorporates both the magnitude of the force and the component of that force along the direction of the displacement. This means that only the portion of the force that acts in the direction of the displacement contributes to the work done.

Transcript

00:11 In this question we have displacement d equals to 1 .5 meter and the given diagram for the block is something like this.

00:23 Okay, so this is our row which is at an angle 30 degree with the horizontal.

00:31 It has tension t so its component along this direction will be t cost theta, t cost theta that is 30 degree and along this direction it will have t sign 30 degree.

00:46 Okay we have weight w along this and normal along this and it will have a friction force that will act backward that is small f okay now for the part a of the problem the work done due to the gravity first of all the work done at w is given by f d cos theta okay so so now we will solve part a.

01:17 So for the work done due to the gravity, wg, we have force as weight w and displacement d but the angle between work done and displacement.

01:31 Displacement is along this direction.

01:34 So the angle between these two will be 90 degree.

01:39 So we will write cos 90 and cos 90 is equal to 0.

01:42 So work done by the gravity is equal to zero newtok so this is the answer for the part a of the problem okay now moving to the part b in which we have to determine the work done by the tension force so the tension force along the displacement is t cost 30 degrees so force will be t cost 30 degree and displacement is d and angle between both is zero so the tension force is 5 .5 newton and cost 30 degree multiplied by d is 1 .5 meter and cost 0 is 1.

02:25 So from here worked on by tension force comes out to be 6 .5 newton.

02:33 Okay? 6 .5 newton.

02:37 Okay so this is the answer for the part b of the problem.

02:42 Now moving to the part c in which we have to determine the work done due to the friction force.

02:48 So the work done due to friction force will be equals to friction force multiplied by displacement, multiplied by cos theta and theta between friction force and displacement will be 180 as both are opposed to each other.

03:02 And here the friction force will be 1 newton and displacement is 1 .5.

03:11 And cos 180 degree...

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