Question

(a) a block with a mass m is pulled along a horizontal surface for a distance x by a constant force f at an angle θ with respect to the horizontal. the coefficient of kinetic friction between block and table is μk. is the force exerted by friction equal to μkmg? if not, what is the force exerted by friction? (assume θ is measured above the horizontal.)
score: 1 out of 1
comment:
(b) how much work is done by the friction force and by f? (dont forget the signs. use the following as necessary: f for the magnitude of f, g, m, x, θ and μk.)
wf = -μk(mg - f - sin(f)x)
wf = fx cos
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(c) identify all the forces that do no work on the block. (select all that apply.)
□ mg
□ friction f
□ f cos(θ)
□ f sin(θ)

Explanation:

Step1: Analyze normal - force

The normal - force $N$ on the block is determined by the vertical - force equilibrium. The vertical component of the applied force is $F\sin\theta$ acting upwards and the weight of the block is $mg$ acting downwards. So, $N + F\sin\theta=mg$, and $N = mg - F\sin\theta$.

Step2: Calculate frictional force

The kinetic frictional force $f_k=\mu_kN$. Substituting $N = mg - F\sin\theta$ into the formula for $f_k$, we get $f_k=\mu_k(mg - F\sin\theta)$.

Step3: Calculate work done by friction

The work - done formula is $W = Fd\cos\theta$. The frictional force $f_k$ acts in the opposite direction of the displacement $x$. So, the work done by the frictional force $W_{f_k}=-f_kx=-\mu_k(mg - F\sin\theta)x$.

Step4: Calculate work done by the applied force

The horizontal component of the applied force is $F\cos\theta$. The work done by the applied force $W_F = Fx\cos\theta$.

Step5: Determine forces doing no work

Work is done when a force has a component in the direction of displacement. The weight $mg$ acts vertically downwards and the displacement is horizontal, so the angle between them is $90^{\circ}$. Using the work formula $W = Fd\cos\theta$, with $\theta = 90^{\circ}$ and $\cos90^{\circ}=0$, the work done by $mg$ is $0$. The vertical component of the applied force $F\sin\theta$ acts perpendicular to the horizontal displacement, so the work done by $F\sin\theta$ is also $0$.

Answer:

(a) No, the force exerted by friction is $f_k=\mu_k(mg - F\sin\theta)$.
(b) Work done by friction: $W_{f_k}=-\mu_k(mg - F\sin\theta)x$; Work done by $\vec{F}$: $W_F = Fx\cos\theta$.
(c) $mg$, $\vec{F}\sin\theta$