Questions: My 6 kg cat wants to push his box of kitty chow off the table so it will spill a snack for his afternoon "tea". The box has a mass of 1 kg (that's 2.2 lb or 9.8 N of gravitational downward force) and the coefficient of friction with the table top for both kitty paws and box is 0.25 . About how much force does he have to push with to get the box to begin to slide? Pick the closest answer to what you calculate using your understanding of static frictional forces. (A) 15 N (B) 25 N (C) 2.5 N (D) He will slide before the box does (E) 2.2 lb (F) 9.8 N Continue Last saved 4:59:46 PM

My 6 kg cat wants to push his box of kitty chow off the table so it will spill a snack for his afternoon "tea". The box has a mass of 1 kg (that's 2.2 lb or 9.8 N of gravitational downward force) and the coefficient of friction with the table top for both kitty paws and box is 0.25 . About how much force does he have to push with to get the box to begin to slide? Pick the closest answer to what you calculate using your understanding of static frictional forces. (A) 15 N (B) 25 N (C) 2.5 N (D) He will slide before the box does (E) 2.2 lb (F) 9.8 N Continue Last saved 4:59:46 PM
Transcript text: My 6 kg cat wants to push his box of kitty chow off the table so it will spill a snack for his afternoon "tea". The box has a mass of 1 kg (that's 2.2 lb or 9.8 N of gravitational downward force) and the coefficient of friction with the table top for both kitty paws and box is 0.25 . About how much force does he have to push with to get the box to begin to slide? Pick the closest answer to what you calculate using your understanding of static frictional forces. (A) 15 N (B) 25 N (C) 2.5 N (D) He will slide before the box does (E) 2.2 lb (F) 9.8 N Continue Last saved 4:59:46 PM
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Solution

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Solution Steps

Step 1: Understanding the Problem

We need to determine the force required for a 6 kg cat to push a 1 kg box so that it begins to slide on a table. The coefficient of static friction between the box and the table is given as 0.25.

Step 2: Calculate the Normal Force

The normal force (\( F_N \)) is the force exerted by the table on the box, which is equal to the gravitational force acting on the box. This can be calculated using the formula: \[ F_N = m \cdot g \] where \( m \) is the mass of the box (1 kg) and \( g \) is the acceleration due to gravity (9.8 m/s\(^2\)).

\[ F_N = 1 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 9.8 \, \text{N} \]

Step 3: Calculate the Maximum Static Friction Force

The maximum static friction force (\( F_{\text{friction}} \)) can be calculated using the coefficient of static friction (\( \mu \)) and the normal force (\( F_N \)): \[ F_{\text{friction}} = \mu \cdot F_N \] \[ F_{\text{friction}} = 0.25 \times 9.8 \, \text{N} = 2.45 \, \text{N} \]

Step 4: Determine the Required Force to Overcome Static Friction

The force required to get the box to begin to slide is equal to the maximum static friction force. Therefore, the cat needs to exert a force of 2.45 N.

Final Answer

The closest answer to 2.45 N is: \[ \boxed{2.5 \, \text{N}} \]

Thus, the answer is (C) 2.5 N.

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