You want to use a rope to pull a 10-kg box of books up a plane inclined 30^∘ above the horizonta

You want to use a rope to pull a 10-kg box of books up a...

You want to use a rope to pull a $10-\mathrm{kg}$ box of books up a plane inclined $30^{\circ}$ above the horizontal. The coefficient of kinetic friction is 0.30 . What force do you need to exert on the other end of the rope if you want to pull the box (a) at constant speed and (b) with a constant acceleration of $0.50 \mathrm{m} / \mathrm{s}^{2}$ up the plane? The rope pulls parallel to the incline.

Key Concepts

Constant Speed vs. Constant Acceleration

For an object moving at constant speed, the net force along the direction of motion is zero, meaning all forces, including friction and gravitational components, are balanced. In contrast, constant acceleration requires a net force equal to the mass multiplied by the acceleration, necessitating an extra force component to achieve the acceleration in addition to counteracting friction and gravitational pull.

Kinetic Friction and the Coefficient of Kinetic Friction

Kinetic friction acts opposite to the direction of motion when an object slides over a surface. It is calculated by multiplying the normal force by the coefficient of kinetic friction. Understanding this concept is key to determining the additional force needed to overcome friction when moving an object along an inclined surface.

Newton's Second Law of Motion

Newton's second law relates the net force acting on an object to its mass and the acceleration it experiences. This principle is foundational for setting up equations of motion in dynamics, particularly when calculating forces required for constant speed (zero net force) or for a specified acceleration.

Decomposition of Forces on an Inclined Plane

When analyzing an object on an inclined plane, it is essential to decompose the gravitational force into components parallel and perpendicular to the incline. The parallel component drives the motion down the plane while the perpendicular component determines the normal force, which is crucial in assessing frictional forces.

Transcript

00:02 This question, the diagram would look like this one that there is a block word there having mass and f which pulling angle of this inclination is 30 degree which is also given and mass which is equal to 10 kg given over there nu k which is given also which is equal to 0 .3 theta is 30 degree so we don't need to write again and again and obviously we have we have to resolve the forces over the there will be m g force there will be this angle also will be theta which is equals to 30 degree there will be this force as well and component of m g actually in this direction and there will be a normal force as well so m g mg cost theta will be in this direction mg sign theta theta over there is 90 degree that is 30 degree so we can say that normal force over there and there is no motion is normal it's just mg cost 30 degree and the friction force f we already know will be nothing but mu k times of normal which is also equals to mu k times of m g cos of 30 degree so first of all at constant speed acceleration will be zero so we can say that acceleration will be zero for that f minus of friction force plus mg sine theta which is also acting up to the motion m g sign of 30 degree equals to 0 which means f equals to friction force plus m g sign 30 degree which equals to sign 30 degree we can also use 1 by 2 in plus of science 30 degree and f over the friction force which we already calculated acceleration mu k times of m g cos of 30 degree plus m g sign of 30 degree so when we put all this value 0 .3 into mass 10 into 9 .81 cost 30 degree 3 upon 2 plus same 10 into 9 .81 into 1 upon 2 so f over there that comes out to be 7454 newton and in the next part constant acceleration a over there which is equal to 0 .5 is given so so, acceleration a over there is 0 .5 meter per second square.

02:48 So we can say that over there, f is the net pulling force minus of friction force minus plus m g sign 30 degree equals to just ma or mass times of acceleration, which is equals to 0 .5 as well...

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