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9. A 10 kg block of ice slides down a ramp 20 m long,inclined at 10^circ to the horizontal. a) If the ramp is frictionless, what is the acceleration of the block of ice? b) If the coefficient of kinetic friction is 0.10, how long will it take the block to slide down the ramp, if it starts from rest?

Question

9. A 10 kg block of ice slides down a ramp 20 m long,inclined at 10^circ to the horizontal. a) If the ramp is frictionless, what is the acceleration of the block of ice? b) If the coefficient of kinetic friction is 0.10, how long will it take the block to slide down the ramp, if it starts from rest?

9. A 10 kg block of ice slides down a ramp 20 m long,inclined at 10^circ  to the horizontal.
a) If the ramp is frictionless, what is the acceleration of the block of ice?
b) If the coefficient of kinetic friction is 0.10, how long will it take the block to slide down
the ramp, if it starts from rest?

Solution

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UlisesMaster · Tutor for 5 years

Answer

**a) Frictionless Ramp:**<br /><br />* **Forces:** The only force acting on the block along the ramp is the component of gravity parallel to the ramp. This is given by *mgsinθ*, where *m* is the mass, *g* is the acceleration due to gravity (9.8 m/s²), and *θ* is the angle of inclination.<br /><br />* **Newton's Second Law:** We can apply Newton's second law (F = ma) along the ramp:<br /> *mgsinθ = ma*<br /><br />* **Solving for Acceleration:** The mass *m* cancels out, leaving:<br /> *a = gsinθ*<br /> *a = (9.8 m/s²) * sin(10°)*<br /> *a ≈ 1.70 m/s²*<br /><br />**b) Ramp with Friction:**<br /><br />* **Forces:** Now there are two forces acting along the ramp: the component of gravity parallel to the ramp (*mgsinθ*) and the force of kinetic friction (*fₖ*). The force of kinetic friction is given by *fₖ = μₖN*, where *μₖ* is the coefficient of kinetic friction and *N* is the normal force. The normal force is equal to the component of gravity perpendicular to the ramp, which is *mgcosθ*. Therefore, *fₖ = μₖmgcosθ*.<br /><br />* **Newton's Second Law:** Applying Newton's second law along the ramp:<br /> *mgsinθ - fₖ = ma*<br /> *mgsinθ - μₖmgcosθ = ma*<br /><br />* **Solving for Acceleration:** Again, the mass *m* cancels out:<br /> *a = gsinθ - μₖgcosθ*<br /> *a = g(sinθ - μₖcosθ)*<br /> *a = (9.8 m/s²)[sin(10°) - (0.10)cos(10°)]*<br /> *a ≈ 0.74 m/s²*<br /><br />* **Kinematics:** We can use the following kinematic equation to find the time it takes to slide down the ramp:<br /> *d = v₀t + (1/2)at²*<br /> where *d* is the distance (20 m), *v₀* is the initial velocity (0 m/s since it starts from rest), *a* is the acceleration (0.74 m/s²), and *t* is the time.<br /><br />* **Solving for Time:** Since *v₀ = 0*, the equation simplifies to:<br /> *d = (1/2)at²*<br /> *t = sqrt(2d/a)*<br /> *t = sqrt(2 * 20 m / 0.74 m/s²)*<br /> *t ≈ 7.35 s*<br /><br /><br />**Answers:**<br /><br />a) The acceleration of the block on the frictionless ramp is approximately **1.70 m/s²**.<br /><br />b) The time it takes the block to slide down the ramp with friction is approximately **7.35 s**.<br />
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