To determine the distance it takes for the sled to stop, we can start by calculating the force of friction acting on the sled and then use Newton's second law to find the acceleration (or deceleration) due to friction.
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Calculate the normal force (N): The normal force acting on the sled is equal to its weight when on a flat surface. Since the sled is on a hill, the normal force is given by:
\[ N = m \cdot g \]
where:
- \( m = 200 , \text{kg} \) (mass of the sled)
- \( g = 10 , \text{m/s}^2 \) (acceleration due to gravity)
So,
\[ N = 200 , \text{kg} \cdot 10 , \text{m/s}^2 = 2000 , \text{N} \]
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Calculate the frictional force (f_friction): The frictional force can be calculated using the coefficient of friction (μ) and the normal force:
\[ f_{\text{friction}} = \mu \cdot N \]
where:
- \( \mu = 0.02 \)
Thus,
\[ f_{\text{friction}} = 0.02 \cdot 2000 , \text{N} = 40 , \text{N} \]
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Calculate the deceleration (a): By Newton's second law, the net force (which is just the frictional force in this case) equals mass times acceleration:
\[ f_{\text{friction}} = m \cdot a \]
Rearranging, we find:
\[ a = \frac{f_{\text{friction}}}{m} = \frac{40 , \text{N}}{200 , \text{kg}} = 0.2 , \text{m/s}^2 \]
Since friction is acting to slow down the sled, this acceleration is negative:
\[ a = -0.2 , \text{m/s}^2 \]
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Find the distance to stop (d): We can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance:
\[ v^2 = u^2 + 2ad \]
where:
- \( v = 0 , \text{m/s} \) (final velocity when it stops)
- \( u = 6 , \text{m/s} \) (initial velocity)
- \( a = -0.2 , \text{m/s}^2 \) (deceleration)
- \( d \) is the distance we are trying to find.
Plugging in these values:
\[ 0 = (6 , \text{m/s})^2 + 2(-0.2 , \text{m/s}^2)d \]
This simplifies to:
\[ 0 = 36 - 0.4d \]
Rearranging gives:
\[ 0.4d = 36 \]
\[ d = \frac{36}{0.4} = 90 , \text{m} \]
Thus, the sled will take 90 meters to stop.
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