(a) A block with a mass m is pulled along a horizontal...

(a) A block with a mass $m$ is pulled along a horizontal surface for a distance $x$ by a constant force $\overrightarrow{\mathbf{F}}$ at an angle $\theta$ with respect to the horizontal. The coefficient of kinetic friction between block and table is $\mu_{k}$. Is the force cxerted by friction equal to $\mu_{k} m g ?$ If not, what is the force exerted by friction? (b) How much work is done by the fricuon force and by $\overrightarrow{\mathbf{F}}$ (Don't forget the signs.) (c) Identify all the forces that do no work on the block. (d) Let $m=2.00 \mathrm{~kg}, x=4.00 \mathrm{~m}, \theta=37.0^{\circ}, F=15.0 \mathrm{~N}$, and $\mu_{k}=$ $0.400$, and find the answers to parts (a) and (b).

(a) A block with a mass $m$ is pulled along a horizontal surface for a distance $x$ by a constant force $\overrightarrow{\mathbf{F}}$ at an angle $\theta$ with respect to the horizontal. The coefficient of kinetic friction between block and table is $\mu_{k}$. Is the force cxerted by friction equal to $\mu_{k} m g ?$ If not, what is the force exerted by friction? (b) How much work is done by the fricuon force and by $\overrightarrow{\mathbf{F}}$ (Don't forget the signs.) (c) Identify all the forces that do no work on the block. (d) Let $m=2.00 \mathrm{~kg}, x=4.00 \mathrm{~m}, \theta=37.0^{\circ}, F=15.0 \mathrm{~N}$, and $\mu_{k}=$
$0.400$, and find the answers to parts (a) and (b).

Key Concepts

Forces That Do No Work

Certain forces, such as the normal force and gravitational force in scenarios where there is no displacement in their direction, can do no work on an object. Recognizing which forces do no work is critical for simplifying the analysis of energy transfer in a system.

Normal Force

The normal force is the perpendicular contact force exerted by a surface on an object, which balances the object's weight and any other vertical components of applied forces. When additional forces act vertically, such as the vertical component of an applied force, the normal force is adjusted accordingly, affecting the calculation of frictional forces.

Work

Work is defined as the energy transferred by a force acting over a distance in the direction of the force. Calculating work involves the dot product of force and displacement vectors, which accounts for both the magnitude and the direction, including the proper sign that indicates whether energy is added to or removed from the system.

Kinetic Friction

Kinetic friction is the resistive force that opposes the relative motion between two contacting surfaces that are sliding against each other. Unlike static friction, kinetic friction has a constant magnitude for given surfaces and is modeled as the product of the coefficient of kinetic friction and the normal force between the surfaces.

Force Decomposition

When a force is applied at an angle, it is essential to break it down into its horizontal and vertical components. This process allows for the separate analysis of the effects each component has on the motion of an object, such as contributing to displacement in the horizontal direction or affecting the normal force acting on the object.

Transcript

00:01 So here we have a quick free body diagram of the system, and we have our normal force, gravitational force of friction, and our applied force applied at an angle theta made with the horizontal.

00:12 We can say that in this case, we have this mass is moving.

00:18 So for part a, we can say that then the kinetic frictional force would be equal to the coefficient of kinetic friction multiplied by the normal force.

00:27 In this case, we can say that the sum of forces in the y direction would be equal to the mass times the acceleration of the y direction.

00:35 We have translational equilibrium in the y direction, so this will be equal to zero.

00:40 And we essentially have the normal force plus the force applied sign of theta, so the y component of that applied force minus mg equaling zero.

00:52 And so the normal force n would be equaling to m g minus f sine of theta therefore the coefficient rather the magnitude of the kinetic frictional force would be equal to mu sub k the coefficient of kinetic friction multiplied by m g minus f sine of theta for part b then the work done by the frictional force would be equaling to the magnitude of the frictional force.

01:29 We can say negative mu sub k, m g minus f sine of theta.

01:35 The negative is coming from the fact that the frictional force is acting in the opposite direction as the opposite direction as the displacement.

01:46 And so if the force is acting in the opposite direction or anti -anty parallel to, the displacement force, the work being done is going to be negative.

02:00 And then this would be multiplied by x, the displacement in the x direction.

02:06 And then the work done by the applied force, we can simply say w would be the x component of the applied force f cosine of theta multiplied by the displacement x.

02:21 And we can say that for part c, the work done by gravity equals the work done, by the normal force equals the work done by the y component of the applied force, equaling of course, zero joules.

02:37 And that is because the normal force vector is perpendicular to the displacement vector.

02:45 Same goes for the gravitational force being perpendicular to the displacement vector.

02:52 And of course, the y component of the applied force, of course being perpendicular to the displacement in the x direction...

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