Question

type 3. two - dimensional dynamics on an incline plane - including frictional forces: for this block on the incline assume that it starts at rest. dont forget to draw the free body diagram! calculate: a) the net force on the mass, b) the acceleration of the mass, and c) velocity at the bottom of the ramp. a 6.7 kg block on an incline that is 2.6 m long; the coefficient of friction on the block is 0.53. a) fnet = f|| - ff f||=(6.7 kg)(9.81)sin41 - 6.7(9.81)(0.53) =43.1 - 34.8 =8.3 n fnet = 8.3 n b) a = fnet/m = 8.3 n/6.7 kg = 1.24 m/s²

Explanation:

Step1: Calculate the gravitational - force component along the incline

The gravitational - force is $F_g = mg$, where $m = 6.7\ kg$ and $g = 9.8\ m/s^2$. The component of the gravitational - force along the incline is $F_{g\parallel}=mg\sin\theta$, with $\theta = 41^{\circ}$. So $F_{g\parallel}=6.7\times9.8\times\sin41^{\circ}\approx6.7\times9.8\times0.656\approx43.1\ N$.

Step2: Calculate the normal force

The normal force $N$ on the block on the incline is $N = mg\cos\theta$. So $N = 6.7\times9.8\times\cos41^{\circ}\approx6.7\times9.8\times0.755\approx49.5\ N$.

Step3: Calculate the frictional force

The frictional force $F_f=\mu N$, where $\mu = 0.53$. So $F_f = 0.53\times49.5\approx26.2\ N$.

Step4: Calculate the net force

The net force $F_{net}=F_{g\parallel}-F_f$. So $F_{net}=43.1 - 26.2=16.9\ N$.

Step5: Calculate the acceleration

According to Newton's second - law $F = ma$, so $a=\frac{F_{net}}{m}$. Substituting $F_{net}=16.9\ N$ and $m = 6.7\ kg$, we get $a=\frac{16.9}{6.7}\approx2.52\ m/s^2$.

Step6: Calculate the velocity at the bottom of the ramp

We use the kinematic equation $v^2 = v_0^2+2ax$, where $v_0 = 0\ m/s$ (starts from rest), $a = 2.52\ m/s^2$, and $x = 2.6\ m$. Then $v^2=0 + 2\times2.52\times2.6=13.104$, and $v=\sqrt{13.104}\approx3.62\ m/s$.

a)

The net force $F_{net}=16.9\ N$.

b)

The acceleration $a\approx2.52\ m/s^2$.

c)

The velocity at the bottom of the ramp $v\approx3.62\ m/s$.

Answer:

Step1: Calculate the gravitational - force component along the incline

The gravitational - force is $F_g = mg$, where $m = 6.7\ kg$ and $g = 9.8\ m/s^2$. The component of the gravitational - force along the incline is $F_{g\parallel}=mg\sin\theta$, with $\theta = 41^{\circ}$. So $F_{g\parallel}=6.7\times9.8\times\sin41^{\circ}\approx6.7\times9.8\times0.656\approx43.1\ N$.

Step2: Calculate the normal force

The normal force $N$ on the block on the incline is $N = mg\cos\theta$. So $N = 6.7\times9.8\times\cos41^{\circ}\approx6.7\times9.8\times0.755\approx49.5\ N$.

Step3: Calculate the frictional force

The frictional force $F_f=\mu N$, where $\mu = 0.53$. So $F_f = 0.53\times49.5\approx26.2\ N$.

Step4: Calculate the net force

The net force $F_{net}=F_{g\parallel}-F_f$. So $F_{net}=43.1 - 26.2=16.9\ N$.

Step5: Calculate the acceleration

According to Newton's second - law $F = ma$, so $a=\frac{F_{net}}{m}$. Substituting $F_{net}=16.9\ N$ and $m = 6.7\ kg$, we get $a=\frac{16.9}{6.7}\approx2.52\ m/s^2$.

Step6: Calculate the velocity at the bottom of the ramp

We use the kinematic equation $v^2 = v_0^2+2ax$, where $v_0 = 0\ m/s$ (starts from rest), $a = 2.52\ m/s^2$, and $x = 2.6\ m$. Then $v^2=0 + 2\times2.52\times2.6=13.104$, and $v=\sqrt{13.104}\approx3.62\ m/s$.

a)

The net force $F_{net}=16.9\ N$.

b)

The acceleration $a\approx2.52\ m/s^2$.

c)

The velocity at the bottom of the ramp $v\approx3.62\ m/s$.